Question

2C₄H₁₀+13O_{2 ----->} 8 CO₂+10 H₂O

If there is excess C₄H₁₀ how many grams of O₂ are needed to produce 144 rams of CO₂. ( Show your work)

Answer 1

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = 144 g

mol of CO2 = (mass)/(molar mass)

= 144/44.01

= 3.272 mol

we have the Balanced chemical equation as:

2C4H10+13O2 -----> 8 CO2+10 H2O

From balanced chemical reaction, we see that

when 8 mol of CO2 reacts, 13 mol of O2 is required

mol of O2 required = (13/8)* moles of CO2

= (13/8)*3.272

= 5.317 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 5.317*32

= 170 g

Answer: 170 g