MOLE CONCEPT
The molar mass of H2SO4 is ...
Ar H=1, S=32, O=16
H2SO4
Mr = 2 x Ar H + 1 x Ar S + 4 x Ar O
Mr = 2(1) + 32 + 4(16) =98
The balanced equation for the following reaction :
Na2CO3 + HCl -> NaCl + H2O + CO2
is...
Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2
Find the mole of CH3COOH dissolved in 100 ml solution of CH3COOH 0.04 M.
n = C x V
n = 0.04 x 0.1 = 0.004 mol
Given a percent yield of a reaction is 75%, and the mass of the actual yield is 8 grams. Find the mass of the theoretical yield!
Mass of theoretical yield = Mass of actual yield: the percent yield
= 8: 0.75 = 10.67 grams.
A 7 mol of CH4 was combusted with 10 mol of O2 via the following reaction :
CH4 + 2 O2 -> CO2 + 2 H2O
The limiting reactant is ...
CH4 = 7 :1 = 7
O2 = 10 : 2 = 5
O2 < CH4
The LR is O2
Find the Mr of hydrated copper (II) sulfate,
CuSO4.5 H2O
Ar Cu =63.5, S=32, O=16,H=1
Mr of CuSO4.5 H2O
Mr CuSO4 + 5 x Mr H2O
Mr CuSO4 = 63.5 + 32 + 4(16) =159.5
Mr H2O = 2(1) + 16 = 18
Mr of CuSO4.5 H2O = Mr CuSO4 + 5 x Mr H2O
Mr of CuSO4.5 H2O = 159.5 + 5(18) = 249.5
Given the following reaction :
Mg(s) + 2 HCl (aq) -> MgCl2 (aq) + H2 (g)
Find the mole of H2 formed when 2 mol of Mg is used in the reaction.
Mole ratio of Mg : HCl : MgCl2 : H2 = 1 : 2: 1 : 1
Mole of Mg = 2 mol
Mole of H2 = 2 mol
Find the mass of 2 mole NH3 if Ar N =14, H=1
Mr NH3 = 1 x Ar N + 3 x Ar H
Mr NH3 = 14 + 3 (1) = 17
Mass NH3 = mol x Mr
Mass NH3 = 2 x 17 = 34 gram
Given the following reaction :
2 SO2 (g) + O2 (g) -> 2 SO3 (g)
Find the mass of SO3 formed when 6.4 grams of SO2 are used in the reaction!
Ar S= 32, O=16
Mole ratio SO2 : O2 : SO3 = 2: 1: 2
Mr SO2 = 1(32) + 2(16) =64
Mole of SO2 = mass : Mr = 6.4 : 64 = 0.1 mol
Mole of SO3 = 0.1 mol
Mr SO3 = 1(32) + 3(16) = 80
Mass SO3 = mole x Mr = 0.1 x 80 = 8 gram.
Find the mass of glucose dissolved in 200 ml of the glucose solution 0.06 M. Mr glucose = 180
mole of glucose = C x V = 0.06 x 0.2 = 0.012 mol
Mass of glucose = mole x Mr = 0.012 x 180 = 2.16 grams.
Given a 100 ml AgNO3 solution 0.05 M, it is reacted with HCl via the following reaction :
AgNO3 + HCl -> AgCl +HNO3
The mass of AgCl formed is 0.6567 grams. Find the percent yield if the Mr AgCl = 143.5
Mole of AgNO3 = C x V = 0.05 x 0.1 = 0.005 mol
Mole of AgCl = 0.005 mol
Theoretical yield of AgCl = mole x Mr = 0.005 x 145.5 = 0.7175 grams
Percent yield = ( 0.6567 : 0.7175) x 100% = 91.52 %
A 7 mol of CH4 was combusted with 10 mol of O2 via the following reaction :
CH4 + 2 O2 -> CO2 + 2 H2O
The final mole of CH4 is ...
Initial mole of CH4 = 7 mol
LR = O2
Mole of CH4 used in the reaction = 5 mol
Final mole of O2 = 7 mol - 5 mol = 2 mol
Find the volume of 1.5 mol CO2(g) at RTP.
Vm at RTP = 24.4 L
Volume of CO2 at RTP = mol x Vm
V CO2 = 1.5 x 24.4 L = 36.6 L
Given the following reaction :
Mg (s) + 2 HCl (aq) -> MgCl2 (aq) + H2 (g)
Find the volume of H2 formed when 0.4 mole of HCl is reacted at P, T.
The 1 ml of N2 mass is 7 mg at P, T.
Ar N=14
Mole of HCl = 0.4 mol
Mole of H2 = 0.4 : 2 = 0.2 mol = 200 mmol
At P, T (compare the gas if H2 and N2)
Mr N2 = 2(14) =28
Mole of N2 = 7: 28 = 0.25 mmol
V1 / n1 = V2 /n2
1 ml/ 0.25 mmol = VH2 /200 mmol
V H2 = 800 ml
Find the mass of 3.01 x 10 20 molecules of NH3 if Mr of NH3 = 17.
NA = 6.02 x10 23
Mole of NH3 = (3.01 x 10 20 ) : ( 6.02 x10 23 )= 5 x10-4
Mass = mole x Mr = 0.0005 x 17 = 0.0085 grams.
A 7 mol of CH4 was combusted with 10 mol of O2 via the following reaction :
CH4 + 2 O2 -> CO2 + 2 H2O
Find the mass of CO2 formed! Mr CO2 = 44
LR = O2
The mole of CO2 formed = 10 mol : 2 = 5 mol
Mass CO2 = mol x Mr = 5 x 44 = 220 gram.
Find the mass of 11.2 mL of CH4(g) at STP.
Ar C=12, H=1
Vol CH4 = 11.2 : 1000 = 0.0112 L
Mol CH4 = V : Vm = 0.0112 L : 22.4 L = 0.0005 mol
Mr CH4 = 12 + 4(1) = 16
Mass CH4 = mol x Mr = 0.0005 x 16 = 0.008 gram