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jeka94
3 years ago
10

Suppose a grill lighter contains 50.0 g of butane. How many grams of butane in the lighter would have to be burned to produce 17

.9 L of carbon dioxide at STP
Chemistry
1 answer:
Hitman42 [59]3 years ago
5 0

<span>Answer is: mass of burned butane is 11.6 g.</span>

Chemical reaction: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.

m(butane) = 50,0 g.

<span> V(CO</span>₂) = 17,9 L.<span>
n(CO</span>₂) = V(CO₂) ÷ Vm.<span>
n(CO</span>₂) = 17,9 L ÷ 22,4 L/mol.<span>
n(CO</span>₂) = 0,8 mol.<span>
From chemical reaction n(CO</span>₂) : n(C₄H₁₀) = 8 : 2.<span>
n(C</span>₄H₁₀) = 0,8 mol ÷ 4.<span>
n(C</span>₄H₁₀) = 0,2 mol.<span>
m(C</span>₄H₁₀) = n(C₄H₁₀) · M(C₄H₁₀).<span>
m(C</span>₄H₁₀) = 0,2 mol · 58 g/mol.<span>
m(C</span>₄H₁₀) = 11,6 g.

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