Stoichiometry
What is the percent mass of oxygen in Al(OH)3?
A. 34.59%
B. 3.88%
C. 65.41%
D. 20.51%
E. 61.53%
E!
We can assume a 100g sample of this compound, and the formula mass will need to be used as a multiplier. Additionally, one mole of this compound contains three moles of oxygen:
Which solution contains the FEWEST moles of sodium ions?
A. 50.0 mL of 0.75 M Na2SO4 solution
B. 75.0 mL of 0.50 M Na3PO4 solution
C. 95.0 mL of 0.90 M NaCl solution
D. Both A and B
E. Both B and C
A!
The number of moles of sodium ions can be found by multiplying the volume (in L) by the concentration, and then using the mole ratio in the formulas
This gives 0.075 moles of Na+ in the Na2SO4 solution, 0.1125 moles of Na+ in the Na3PO4 solution, and 0.0855 moles of Na+ in the NaCl solution.
If 3.0 moles of NO and 2.0 moles of O2 are reacted until one of these two reactants is completely consumed, how many moles of which reactant will remain? 2 NO (g) + O2 (g) --> 2 NO2 (g)
A. 0.5 mol O2
B. 0.5 mol NO
C. 1.5 mol O2
D. 1.5 mol NO
A!
Three moles of NO will require 1.5 moles of O2 to react completely, leaving 0.5 moles of O2 unreacted. Two moles of O2 requires 4 moles of NO to react completely, but only 3 moles of NO are available. This makes NO the limiting reactant.
Sucrose is produced in plants by the following reaction:
12CO2 (g) + 11H2O (l) --> C12H22O11 (s) + 12O2(g) ΔH = 5640 kJ
How many moles of carbon dioxide are needed in a reaction that consumed 2.0 X 10^3 kJ of energy?
A. 0.030 moles
B. 2.8 moles
C. 4.3 moles
D. 34 moles
E. 52 moles
C!
According to the balanced equation, 12 moles of CO2 require an input of 5640 kJ of energy to react. Stoichiometry is used to determine how many moles of CO2 are needed in order to require a larger input of energy
Which of the following is balanced correctly?
A. CH4 + 2 O2 → CO2 + 2 H2O
B. 6 H2O + P4O10 → 4 H3PO3
C. 3 H2 + 3 N2 → 2 NH3
D. 2 C3H8O + 3 O2 → 2 CO2 + 4 H2O
A!
A chemical equation is balanced when the same number of atoms are in the reactants and products. The reactants in B contain 16 oxygen atoms, but the products contain only 12 (B). The reactants in C contain 6 nitrogen atoms, but the product contain only 2 (C). The reactants in D contain 6 carbon atoms, 16 hydrogen atoms, and 8 oxygen atoms. The products contain 2 carbon atoms, 8 oxygen atoms, and 8 hydrogen atoms (D).
A compound was found to contain 88.8% carbon by mass with the balance being hydrogen. The molecular mass of this compound is four times the empirical formula mass. What is the compound’s chemical formula?
A. C4H4
B. C4H12
C. C8H8
D. C4H8
E. C8H12
E!
Assuming a 100g sample, this compound will be composed of 88.8g carbon and 11.2g hydrogen. The coefficients in the formula are found by converting these masses into moles and dividing each by the smaller number
Dividing each by 7.39 gives an initial formula of C1H1.5, and whole numbers are obtained by multiplying by a factor of two to give C2H3 as the empirical formula. Because the molecular mass of the compound is four times the empirical formula mass, the formula is multiplied by a factor of four to give C8H12.
What mass of glucose (C6H12O6, 180.2 g/mol) must be used to prepare 500 mL of a solution that is 2.50 M in glucose?
A. 625 g
B. 125 g
C. 450 g
D. 1.25 g
E. 225 g
E!
For solution stoichiometry, a volume is multiplied by the concentration in order to get to moles. Moles can then be converted to mass by multiplying by the molar mass
When 14.0 g of calcium metal is reacted with water, 5.00 g of calcium hydroxide, 74.09 g/mol, is produced. Using the following balanced equation, calculate the percent yield for the reaction.
Ca (s) + 2 H2O (l) → Ca(OH)2 (aq) + H2 (g)
A. 9.67%
B. 19.3%
C. 35.7%
D. 66.0%
B!
In order to calculate the percent yield of this reaction, the theoretical yield must be calculated first. Only the mass of calcium is given, so it is assumed to be the limiting reagent. The given mass is converted to moles, the mole ratio is used to get moles of Ca(OH)2, and the formula mass of Ca(OH)2 is used to get mass of Ca(OH)2:
At 25 ºC and 1 atm the following reaction releases approximately 198.5 kJ of heat while absorbing approximately 2.5 kJ of work at constant pressure. Which of the following is true?
2 SO2 (g) + O2 (g) --> 2 SO3 (g)
A. The reaction is endothermic and ΔE is ‒196.0 kJ and ΔH is ‒198.5 kJ.
B. The reaction is endothermic and ΔE is ‒198.5 kJ and ΔH is ‒196.0 kJ.
C. The reaction is exothermic and ΔE is ‒196.0 kJ and ΔH is ‒198.5 kJ.
D. The reaction is exothermic and ΔE ‒198.5 kJ and ΔH is ‒196.0 kJ.
C!
If the reaction releases 198.5 kJ of heat, q = –198.5 kJ. If it absorbs 2.5 kJ of work, w = +2.5 kJ. DE = q + w DE = –198.5 kJ + 2.5 kJ = –196.0 kJ (B, D)
What is the oxidation number of the iodine atom in the iodite ion, IO2 – ?
A. +4
B. –1
C. –3
D. +1
E. +3
E!
The most common oxidation number for oxygen is –2. With two oxygen atoms in the formula, the total negative charge is equal to –4. Since the charge of the ion is –1, the oxidation number of the iodine has to be +3.
How many grams of N2H2 would be required to react completely with 99.0 g of O2 according to the balanced chemical equation?
2 N2H2 (g) + 5 O2 (g) → 4 NO2 (g) + 2 H2O (g)
A. 37.1 g
B. 42.1 g
C. 74.2 g
D. 232 g
E. 265 g
A!
The chemical equation is used to relate the number of moles of one substance to the number of moles of another. If grams of O2 is given, it needs to be converted to moles. The coefficients are used as mole ratios, and the formula mass of N2H2 is used to get back to grams
In the following reaction involving a weak acid, the spectator ions are:
HNO2 (aq) + KOH (aq) --> KNO2 (aq) + H2O (l)
A. K+ and OH
B. K+ only
C. H+ and OH‒
D. NO2 only
E. H+ and NO2
B!
KOH is a strong base and as such, it will dissociate completely into its ions (K+ and OH– ). This makes HNO2 the weak acid, and weak acids do not dissociate completely. Dissociation is often shown as an equilibrium: HNO2 (aq) ⇌ H+ (aq) + NO2 – (aq) The complete ionic equation is therefore: HNO2 (aq) + K+ (aq) + OH– (aq) ⇌ K+ (aq) + NO2 – (aq) + H2O (l) The spectator ions are those that appear on both sides of the equation. For this reaction, K+ is the only spectator ion.
Hydrogen cyanide is produced from the reaction of gaseous ammonia, oxygen, and methane. If you have 1200 g of each reactant, which would be the limiting reagent for the production of hydrogen cyanide?
2NH3 (g) + 3O2 (g) + 2CH4 (g) -->2HCN (g) + 6H2O (g)
A. NH3
B. O2
C. CH4
D. None of the above
B!
The limiting reactant is the one that is used up first. The chemical equation can be used once the mass is converted to moles. Then the mole ratio between each reactant is used as a multiplier to find out how many moles of HCN can be formed
Given the following reactions:
N2 (g) + O2 (g) → 2 NO (g) ΔH = +180.7 kJ
NO (g) + ½ O2 (g) → NO2 (g) ΔH = –56.6 kJ
What is the enthalpy for the following reaction?
2 NO2 (g) → N2 (g) + 2 O2 (g) ΔH = ?
A. 67.5 kJ
B. 293.9 kJ
C. 124.1 kJ
D. –67.5 kJ
E. –293.9 kJ
D!
The enthalpy of the third reaction can be found by manipulating and combining the first two reactions (Hess’s Law). N2 is a product in the third equation, but it is a reactant in the first reaction. This means the first reaction needs to be flipped, along with the sign of DH: 2 NO (g) → N2 (g) + O2 (g) ΔH = –180.7 kJ O2 is a product in the third reaction, and it appears as a reactant in both of the first two reactions. Flipping the first reaction in the previous step gets O2 in the products. NO2 is a reactant in the third equation, but it is a product in the second reaction. This means the second reaction needs to be flipped, and the sign for DH is changed. It also needs to be doubled since two moles of NO2 are needed in the third reaction: 2 NO2 (g) → 2 NO (g) + O2 (g) ΔH = +113.2 kJ Now, the two equations can be added together to give the third equation:
2 NO (g) → N2 (g) + O2 (g) ΔH = –180.7 kJ
2 NO2 (g) → 2 NO (g) + O2 (g) ΔH = +113.2 kJ
2 NO2 (g) → N2 (g) + 2 O2 (g) DH = –67.5 kJ
Which of the following is an oxidation-reduction reaction?
A. 2 NaCl (aq) + Ca(NO3)2 (aq) → 2 NaNO3 (aq) + CaCl2 (aq)
B. CrCl3 (aq) + 3 KOH (aq) → 3 KCl (aq) + Cr(OH)3 (s)
C. KOH (aq) + HCl (aq) → KCl (aq) + H2O (l)
D. 2 CrCl2 (aq) + CuCl2 (aq) → 2 CrCl3 (aq) + Cu (s)
D!
An oxidation-reduction reaction is one in which electrons are transferred from one or more atoms in the reactants to one or more atoms in the products. The reactants and the products in A are all strong electrolytes (no net ionic equation, A). Two aqueous solutions react in B to form a solid (precipitation, B). A strong acid reacts with a strong base in C to form an ionic salt and water (acid-base, C). Two aqueous solutions react to form a solid in D, but chromium in CrCl2 has an oxidation number of +2. In CrCl3, it has an oxidation number of +3. Copper in CuCl2 has an oxidation number of +2. Solid copper has an oxidation number of 0. Therefore, chromium is oxidized while copper is reduced (an oxidation-reduction reaction).
A compound was found to be, by mass, 50.0% carbon, 5.6% hydrogen with the remainder being oxygen. What is the compound’s empirical formula?
A. C4H6O3
B. CH2O
C. C4H5O3
D. C2H2O
E. C3H4O2
E!
On a percentage basis, we can assume a 100g sample of the compound. This means it contains 50.0g carbon, 5.6g hydrogen, and 44.4g oxygen. Each mass is converted to moles, and the mole ratios are determined by dividing each by the smallest number.
Which of each pair is the stronger electrolyte?
Pair I: C6H12O6 or NH4Cl
Pair II: HI or HF
A. C6H12O6 and HI
B. C6H12O6 and HF
C. NH4Cl and HI
D. NH4Cl and HF
C!
C6H12O6 (i.e. glucose) is a covalent compound that dissolves in water but does not dissociate into any ions. NH4Cl is an ionic compound that completely dissociates into NH4 + and Cl– ions (A, B). HI and HF are both acids, but HI is a strong acid while HF is a weak acid (D).
What mass of water is actually produced by the reaction of 50.0 g CH3OH with an excess of O2 when the yield is 53.2%?
2 CH3OH (l) + 3 O2 (g) --> 2 CO2 (g) + 4 H2O (l)
A. 28.1 g
B. 56.2 g
C. 26.6 g
D. 15.0 g
E. 29.9 g
E!
If O2 is in excess, CH3OH is the limiting reactant, and the mass of H2O formed depends entirely on it. Since a % yield is given (actual / theoretical), the theoretical yield needs to be calculated
Use the given information to calculate the standard heat of formation of methane gas, CH4 (g).
DHf° CO2 (g) = –393.5 kJ/mol
DHf° H2O (g) = –241.8 kJ/mol
DHf° H2O (l) = –285.8 kJ/mol
4H2(g) + CO2(g) → CH4(g) + 2H2O(g) DH°= -164.9kJ
A. –74.8 kJ/mol
B. –316.6 kJ/mol
C. –272.6 kJ/mol
D. 13.2 kJ/mol
A!
The DH° of the reaction can be calculated using the equation DHf° (products) – DHf° (reactants). The coefficients are used as multipliers, and elements in their standard states are not included (H2). –164.9kJ = [DHf° CH4 + 2(DHf° H2O)] – DHf° CO2 –164.9kJ = DHf° CH4 – 483.6kJ –(–393.5kJ) –74.8 kJ/mol = DHf° CH4
In the smelting of iron from iron (III) oxide, according to the next equation, which atom is oxidized?
Fe2O3 (s) + 3 CO (g) --> 2 Fe (s) + 3 CO2 (g)
A. C in CO
B. C in CO2 C.
Fe in Fe2O3
D. O in Fe2O3
E. Fe in Fe metal
A!
An atom will be oxidized if it loses electrons to form a product, so it must be on the reactant side of the chemical equation (B, E). The iron in Fe2O3 has an oxidation number of +3, while elemental iron in the product has an oxidation number of zero. Since it gained electrons, the iron in Fe2O3 has been reduced (C). Oxygen in Fe2O3 has an oxidation number of –2; in CO2, it also has an oxidation number of –2 (D). Carbon in CO has an oxidation number of +2; in CO2, it is +4. Since the carbon in CO lost electrons to form CO2, it has been oxidized.