Stoichiometry
What is stoichiometry?
a branch of chemistry that deals with the application of the laws of definite proportions and of the conservation of mass and energy to chemical activity
what is percent yield
Percent yield is the amount of a compound obtained from a chemical synthesis reaction with respect to the theoretically expected amount. This is a percentage value.
How many grams of PF5 can be formed from 9.46 g of PF3 and 9.42 g of XeF4 in the following reaction?
What is the limiting reagent of
2PF3 + XeF4 ---> 2PF5 + Xe
PF3 ⇒ 9.46 g / 87.968 g/mol = 0.10754 mol
XeF4 ⇒ 9.42 g / 207.282 g/mol = 0.045445 mol
2) Determine limiting reagent:
PF3 ⇒ 0.10754 / 2 = 0.05377
XeF4 ⇒ 0.045445 / 1 = 0.045445
XeF4 is limiting
Carbon monoxide burns in air to produce carbon dioxide according to the following balanced equation: 2 CO(g) + O2(g) 2 CO2(g) + 566 kJ How many grams of carbon monoxide are needed to yield 185 kJ of energy?
18.3 g CO
Some underwater welding is done via the thermite reaction, in which rust (Fe2O3) reacts with aluminum to produce iron and aluminum oxide (Al2O3). In one such reaction, 258 g of aluminum and excess rust produced 464 g of iron. What was the percent yield of the reaction?
464 g = actual yield
Fe2O3 + 2 Al --> Al2O3 + 2 Fe
excess 258 g x g = theoretical yield
87%
Calculate the number of NaBr formula units formed when 50 NBr3 molecules and 57 NaOH formula units react?
2NBr3 + 3NaOH ---> N2 + 3NaBr + 3HOBr
Determine limiting reagent:
NBr3 ⇒ 50 "moles" / 2 = 25
NaOH ⇒ 57 "moles" / 3 = 19
NaOH is the lmiting reagent.
Note that there need be no conversion from grams to moles. Discussions of numbers of molecules uses numbers that are directly proportional to the number of moles and do not need to be converted.
2) Use NaOH : NaBr molar ratio:
3 is to 3 as 57 is to x
x = 57 "moles"
the answer is 57 formula units.
Nitrogen gas combines with oxygen gas according to the following balanced equation: N2(g) + 2 O2(g) + 67.8 kJ 2 NO2(g) Assuming that you have excess nitrogen, how much heat energy must be added to 540 g of oxygen in order to use up all of that oxygen?
572 kJ
Use the balanced equation to find out how many liters of sulfur dioxide are actually produced at STP if 1.5 x 1027 molecules of zinc sulfide are reacted with excess oxygen and the percent yield is 75%. 2 ZnS(s) + 3 O2(g) --> 2 ZnO(s) + 2 SO2(g) 1.5x1027 molecules excess x L = theoretical yield
4.19 10 L SO
C12H22O11 + 12O2 ---> 12CO2 + 11H2O
there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?
1) Calculate moles of sucrose:
10.0 g / 342.2948 g/mol = 0.0292146 mol
2) Calculate moles of oxygen required to react with moles of sucrose:
From the coefficients, we see that 12 moles of oxygen are require for every one mole of sucrose. Therefore:
0.0292146 mol times 12 = 0.3505752 mole of oxygen required
3) Determine limiting reagent:
Oxygen on hand ⇒ 10.0 g / 31.9988 g/mol = 0.3125 mol
Since the oxygen required is greater than that on hand, it will run out before the sucrose. Oxygen is the limiting reagent.
The combustion of propane (C3H8) produces 248 kJ of energy per mole of propane burned. How much heat energy will be released when 1 000 dm3 of propane are burned at STP?
11,071kJ
The Haber process is the conversion of nitrogen and hydrogen at high pressure into ammonia, as follows: 700 g = actual yield
N2(g) + 3 H2(g) --> 2 NH3(g)
x g excess x g = theoretical yield
824 g N2
Suppose 316.0 g aluminum sulfide reacts with 493.0 g of water. What mass of the excess reactant remains?
The unbalanced equation is:
Al2S3 + H2O ---> Al(OH)3 + H2S
Determine moles, then limiting reagent:
Al2S3 ⇒ 316.0 g / 150.159 g/mol = 2.104436 mol
H2O ⇒ 493.0 g / 18.015 g/mol = 27.366 mol
Al2S3 ⇒ 2.104436 / 1 = 2.104436
H2O ⇒ 27.366 / 6 = 4.561
Al2S3 is the limiting reagent.
3) Determine grams of water that react:
The molar ratio to use is 1:6
1 is to 6 as 2.104436 mol is to x
x = 12.626616 mol of water used
12.626616 mol times 18.105 g/mol = 227.4685 g
4) Determine excess:
493.0 g minus 227.46848724 = 265.5 g (to 4 sig figs)
4. Ethyl alcohol burns according to the following balanced equation: C2H5OH(l) + 3 O2(g) 2 CO2(g) + 3 H2O(g) + 1 364 kJ How many molecules of water are produced if 5 000 kJ of heat energy are released?
6.62 x 1024 molecules H2O
. “Slaked lime,” Ca(OH)2, is produced when water reacts with “quick lime,” CaO. If you start with 2 400 g of quick lime, add excess water, and produce 2 060 g of slaked lime, what is the percent yield of the reaction? 2060 g = actual yield
CaO + H2O --> Ca(OH) 2
2400 g excess x g = theoretical yield
65%
Determine just the limiting reagent
How many grams of IF5 would be produced using 44.01 grams of I2O5 and 101.0 grams of BrF3?
6I2O5 + 20BrF3 ---> 12IF5 + 15O2 + 10Br2
Solution:
1) Determine moles:
I2O5 ⇒ 44.01 g / 333.795 g/mol = 0.1318474 mol
BrF3 ⇒ 101.0 g / 136.898 g/mol = 0.7377756 mol
2) Determine limiting reagent:
I2O5 ⇒ 0.1318474 / 6 = 0.02197457
BrF3 ⇒ 0.7377756 / 20 = 0.03688878
I2O5 is limiting.