Question #86eb9

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Question #86eb9

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Answer 1 · 2017-08-04T16:04:44

All hydrocarbons combust to give carbon dioxide and water.......

Explanation:

$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)$ $C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)$

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

And if combust a $40 \cdot g$ $40*g$ mass of propane......you gets.....

$\frac{40 \cdot g}{44.1 \cdot g \cdot m o l^{- 1}} \times 3 \times 44.01 \cdot g \cdot m o l^{- 1} = 119.8 \cdot g \cdot C O_{2}$ $(40*g)/(44.1*g*mol^-1)xx3xx44.01*g*mol^-1=119.8*g*CO_2$

And if combust a $60 \cdot g$ $60*g$ mass of butane......you gets.....

$\frac{60 \cdot g}{58.12 \cdot g \cdot m o l^{- 1}} \times 4 \times 44.01 \cdot g \cdot m o l^{- 1} = 181.7 \cdot g \cdot C O_{2}$ $(60*g)/(58.12*g*mol^-1)xx4xx44.01*g*mol^-1=181.7*g*CO_2$

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Question #86eb9

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