In a butane lighter, 9.7 g of butane combine with with 34.7 g of oxygen to form g g carbon dioxide and 29.3 how many grams of water?

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In a butane lighter, 9.7 g of butane combine with with 34.7 g of oxygen to form g g carbon dioxide and 29.3 how many grams of water?

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Answer 1 · 2017-07-14T07:34:59

Approx. $15 \cdot g$ $15*g$ ......

Explanation:

We need a stoichiometric equation........

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l)$ $C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(l)$

And thus $moles of butane = \frac{9.7 \cdot g}{58.12 \cdot g \cdot m o l^{- 1}} = 0.167 \cdot m o l$ $"moles of butane"=(9.7*g)/(58.12*g*mol^-1)=0.167*mol$ .

And so upon complete combustion we gets......

$0.167 \cdot m o l \times 5 \times 18.011 \cdot g \cdot m o l^{- 1} = 15.0 \cdot g$ $0.167*molxx5xx18.011*g*mol^-1=15.0*g$ water.......

We could have avoided this rigmarole by noting that mass is always with conserved in a chemical reaction. By the problem's specification, we had $(34.7 + 9.7) \cdot g = 44.4 \cdot g$ $(34.7+9.7)*g=44.4*g$ of reactant; we got a mass $29.3 \cdot g$ $29.3*g$ carbon dioxide, and thus there was a balance of $44.4 - 29.3 \cdot g = 15.0 \cdot g$ $44.4-29.3*g=15.0*g$ as required........WHICH MUST REPRESENT THE MASS OF WATER...........

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In a butane lighter, 9.7 g of butane combine with with 34.7 g of oxygen to form g g carbon dioxide and 29.3 how many grams of water?

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