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 data:

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