Question

What volume of carbon dioxide will be produced when 12.0 grams of butane, C4H10(molar mass = 58.15 g/mol), completely burns in oxygen at STP?

Answer 1

Given:

mass of butane = 12g

molar mass of butane = 58.15g

reaction = $2C_{4}H_{10} + 13 O_{2}$  → $8C O_{2} + 10H_{2}O$

To Find:

the volume of carbon dioxide produced =?

Solution:

we know for the given reaction Oxygen ($O_{2}$) is the limiting reagent

Thus, 2mol butane reacts 13 mol oxygen

         12g butane gives   $\dfrac{13 \times 12}{2 \times 58.15}$ mol oxygen

= 1.314 mol oxygen

now, 13 mol oxygen react with 8 mol carbon dioxide

1.314 mol oxygen react to give $\dfrac{8\times 1.314}{13}$ mol carbon dioxide

now the volume of gas at STP = 22.4 L

1.314 × 22.4 L volume of oxygen gas react to give 0.808 × 22.4L carbon dioxide

= 18.09 L carbon dioxide will be produced

Answer 2

Answer:

18.09

Explanation:

chemistry