Propane and oxygen react according to the equation: $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (g)$
If two moles of $C_{3} H_{8} (g)$ is used in the above reaction then determine how many grams of water will produce after the complete combustion?

144.0

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82.0

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8.0

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44.8

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22.4

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Solution

The correct option is A $144.0$
In above reaction $1 m o l e$ of $C_{3} H_{8}$ gives $4 m o l e s$ of $H_{2} 0$
so $2 m o l e s$ of $C_{3} H_{8}$ will give $8 m o l e s$ of $H_{2} 0$
molecular weight of $H_{2} O$ $=$ $18 g r a m / m o l e$
so total weight of produced water $=$ $18 * 8$ $=$ $144 g r a m$

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Similar questions

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

C_{3} H_{8} (g)

C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (g)

C O_{2}

Complete and balance the given reaction:

$C_{3} H_{8} (g) + O_{2} (g) \to$

C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)

44.0

C O_{2} (g)

H_{2} O (I)

C_{3} H_{8} (g)

C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)

Δ H_{r x n}^{\circ} = - 2221.6 k J

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