Amount of water produced by combustion of $3.2$ g of methane in the presence of excess of oxygen is:

5

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3.6

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7.2

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4.2

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Solution

The correct option is D $3.6$ g
$C H_{4} + 2 O_{2} ⟶ C O_{2} + 2 H_{2} O$

$1 : 2$ $1 : 2$

Weight of $C H_{4} = 32 g$

Moles of $C H_{4} = \frac{3.2}{32} = 0.1 m o l e$

$C H_{4} + 2 O_{2} ⟶ C O_{2} + 2 H_{2} O$
$1 : 2$

$0.1$ $1 :$ $2$

$0.1 \times 2 = 0.2 m o l e o f H_{2} O$

Weight of $H_{2} O = m o l e s \times M o l . w t . = 0.2 \times 18 = 3.6 g m$

Amount of $H_{2} O = 3.6 g m$ .

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Amount of water produced by the combustion of 16g of methane is:

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Calculate the amount of water(g) produced by the combustion of 16 gram of Methane.

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