Question

Gaseous butane (CH3(CH2CH3 wll react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 1.7 g of butane is mixed with 1.93 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits. ×1 lg

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)= 1.7 g

number of mol of C4H10,
n = mass of C4H10/molar mass of C4H10
=(1.7 g)/(58.12 g/mol)
= 2.925*10^-2 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 1.93 g

number of mol of O2,
n = mass of O2/molar mass of O2
=(1.93 g)/(32 g/mol)
= 6.031*10^-2 mol
Balanced chemical equation is:
2 C4H10 + 13 O2 ---> 10 H2O + 8 CO2


2 mol of C4H10 reacts with 13 mol of O2
for 0.0292 mol of C4H10, 0.1901 mol of O2 is required
But we have 0.0603 mol of O2

so, O2 is limiting reagent
we will use O2 in further calculation


Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol

According to balanced equation
mol of H2O formed = (10/13)* moles of O2
= (10/13)*0.0603
= 0.0464 mol


mass of H2O = number of mol * molar mass
= 4.639*10^-2*18.02
= 0.836 g

Answer: 0.836 g