According to given reaction, find the mass of $C F_{4}$ formed by the reaction of $8.00 g$ of methane with an excess of fluorine?
$C H_{4} (g) + 4 F_{2} (g) \to C F_{4} (g) + 4 H F (g)$

19 g

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22 g

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38 g

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44 g

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88 g

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Solution

The correct option is B $44 g$
The molar mass of methane is 16 g/mol.
8.00 g methane corresponds to $\frac{8.00 g}{16 g / m o l} = 0.5 m o l$
1 mole of methane gives 1 mole of $C F_{4}$ .
0.5 moles of methane will give 0.5 moles of $C F_{4}$ .
The molar mass of $C F_{4} = 12 + 4 (19) = 88 g / m o l$
0.5 moles $C F_{4} = 0.5 m o l \times 88 g / m o l = 44 g$

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