Question

How many liters of carbon dioxide are produced from the combustion reaction of 165.00 g of butane?

Answer 1

Molar mass of C4H10,
MM = 4*MM(C) + 10*MM(H)
= 4*12.01 + 10*1.008
= 58.12 g/mol


mass(C4H10)= 165.0 g

use:
number of mol of C4H10,
n = mass of C4H10/molar mass of C4H10
=(165.0 g)/(58.12 g/mol)
= 2.839 mol

Balanced chemical equation is:
2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O

According to balanced equation
mol of CO2 formed = (8/2)* moles of C4H10
= (8/2)*2.839
= 11.36 mol

1 mol of gas occupies 22.4 L at STP
So,
volume = number of mol * 22.4 L
= 11.36 mol * 22.4 L/mol
= 254 L

Answer: 254 L