Question #193f6

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Question #193f6

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Answer 1 · 2017-07-01T17:52:41

$C H_{4}$ $CH_4$

Explanation:

Change the grams to moles

$\frac{0.36 g H_{2} O}{18 g / m o l} = 0.02 m o l H_{2} O$ $(0.36 g H_2O)/(18 g//mol) = 0.02 mol H_2O$

$\frac{0.44 g C O_{2}}{44 g / m o l} = 0.01 m o l C O_{2}$ $(0.44 g CO_2)/(44 g//mol) = 0.01 mol CO_2$

The ratio of water to carbon dioxide is 2: 1 so

$C_{x} H_{y} + z O_{2} = 1 C O_{2} + 2 H_{2} O$ $C_xH_y + z O_2 = 1 CO_2 + 2 H_2O$ to balance the equation requires 1 C and 4 H's as well as 2.5 $O_{2}$ $O_2$ so

$1 C H_{4} + 2.5 O_{2} = 1 C O_{2} + 2 H_{2} O$ $1 CH_4 + 2.5 O_2 = 1 CO_2 + 2 H_2O$ or

$2 C H_{4} + 5 O_{2} = 2 C O_{2} + 4 H_{2} O$ $2 CH_4 + 5 O_2 = 2 CO_2 + 4 H_2O$

The compound is methane $C H_{4}$ $CH_4$ .

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Question #193f6

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