Question

grams A to moles A to moles B to grams B 2 C4H10 (l) + 13 O2(g) --> 8 CO2(g) + 10 H2O(g) How many moles of each product will form if 8.12 g of butane (C4H10) reacts completely to form products? Assume there is more than enough of the other reactant. How many grams of CO2 will form?

Answer 1

1)

Molar mass of C4H10,

MM = 4*MM(C) + 10*MM(H)

= 4*12.01 + 10*1.008

= 58.12 g/mol

mass(C4H10)= 8.12 g

number of mol of C4H10,

n = mass of C4H10/molar mass of C4H10

=(8.12 g)/(58.12 g/mol)

= 0.1397 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)*0.1397

= 0.5588 mol

Answer: 0.559 moles of CO2 formed

2)

According to balanced equation

mol of H2O formed = (10/2)* moles of C4H10

= (10/2)*0.1397

= 0.6986 mol

Answer: 0.699 moles of H2O formed

3)

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = number of mol * molar mass

= 0.5588*44.01

= 24.6 g

Answer: 24.6 g of CO2 formed