Answer:
It's a lot of calculation and it took me a bit of time
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EXPLANATIONS
Consider the following combustion reaction: 2 C4H10 + 13 O2 ----> 8 CO2 + 10 H2O. 125 g or C4H10 react with 415 g of O2.
a.) Which mass of CO2 and H2O can be produced?
b.) Which substance is the limiting
a)
Molar mass of C4H10,
MM = 4*MM(C) + 10*MM(H)
= 4*12.01 + 10*1.008
= 58.12 g/mol
mass(C4H10)= 125.0 g
use:
number of mol of C4H10,
n = mass of C4H10/molar mass of C4H10
=(1.25*10^2 g)/(58.12 g/mol)
= 2.151 mol
Molar mass of O2 = 32 g/mol
mass(O2)= 415.0 g
use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(4.15*10^2 g)/(32 g/mol)
= 12.97 mol
Balanced chemical equation is:
2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O
2 mol of C4H10 reacts with 13 mol of O2
for 2.151 mol of C4H10, 13.98 mol of O2 is required
But we have 12.97 mol of O2
so, O2 is limiting reagent
we will use O2 in further calculation
Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol
According to balanced equation
mol of CO2 formed = (8/13)* moles of O2
= (8/13)*12.97
= 7.981 mol
use:
mass of CO2 = number of mol * molar mass
= 7.981*44.01
= 3.512*10^2 g
According to balanced equation
mol of H2O formed = (10/13)* moles of O2
= (10/13)*12.97
= 9.976 mol
use:
mass of H2O = number of mol * molar mass
= 9.976*18.02
= 1.798*10^2 g
Answer:
mass of CO2 = 3.51*10^2 g
mass of H20 = 1.80*10^2 g
b)
O2 is limiting reagent
