$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)$
The combustion of propane, $C_{3} H_{8} (g)$ , proceeds according to the equation above. In the complete combustion of 44.0 grams of propane, determine how many grams of water will be formed.

4.50g

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1.80g

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44.0g

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72.0g

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176g

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Solution

The correct option is D 72.0g
The molar masses of propane and water are 44 g/mol and 18 g/mol respectively.

44.0 grams of propane corresponds to 1 mole of propane.

According to the balance chemical equation, $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)$ ,

1 mole of propane gives 4 moles of water.

Thus, the combustion of 44.0 g (1 mole) of propane will give 4 moles of water.

Mass of water obtained $= 4 \times 18 = 72.0$ g.

Hence, option D is correct.

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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 (g)

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)

44.0

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

C O_{2}

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

Complete and balance the given reaction:

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

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