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Genrish500 [490]

3 years ago

5

2C4H10(g) +13O2(g) → 8CO2(g) +10H2O(g) Suppose a grill lighter contains 50.0 g of butane. How many grams of butane in the lighte

r would have to be burned to produce 17.9 L of carbon dioxide at STP?
A) 46.4 g
B)8.00 g
C)8.95 g
D)11.6 g

1 answer:

MatroZZZ [7]3 years ago

8 0

Answer is: mass of butane is D)11.6 g.
m(butane) = 50,0 g.
V(CO₂) = 17,9 L.
n(CO₂) = V(CO₂) ÷ Vm.
n(CO₂) = 17,9 L ÷ 22,4 L/mol.
n(CO₂) = 0,8 mol.
From chemical reaction n(CO₂) : n(C₄H₁₀) = 8 : 2.
n(C₄H₁₀) = 0,8 mol ÷ 4.
n(C₄H₁₀) = 0,2 mol.
m(C₄H₁₀) = n(C₄H₁₀) · M(C₄H₁₀).
m(C₄H₁₀) = 0,2 mol · 58 g/mol.
m(C₄H₁₀) = 11,6 g.

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