If you start with 36.0mL of C4H10 gas at STP, how many grams are of CO2 can you produce if you have excess oxygen gas?

Mass of CO₂ = 0.352 g

Explanation:

Given dа ta:

Volume of C₄H₁₀ = 36.0 mL (36/1000 = 0.036 L)

Temperature = 273 K

Pressure = 1 atm

Mass of carbon dioxide produced = ?

Solution:

Chemical equation:

2C₄H₁₀ + 13O₂ → 10H₂O + 8CO₂

Number of moles of C₄H₁₀:

PV = nRT

n = PV/RT

n = 1 atm * 0.036 L / 0.0821 atm. L / mol. K * 273 k

n = 0.036 atm. L / 22.4 atm. L / mol

n = 0.002 mol

C₄H₁₀ : CO₂

2 : 8

0.002 : 8/2*0.002 = 0.008 mol

Mass of CO₂:

Mass = number of moles * molar mass

Mass = 0.008 mol * 44 g/mol

Mass = 0.352 g