Question

calculate the amount of h2o produced by the combustion of 4 gram of methane​

Answer 1

Answer:

$\tt CH_{4 \: (g)} + \bigg(x + \dfrac{y}{4} \bigg) O_{2 \: (g)} \longrightarrow x CO_{2\:(g) } + \dfrac{y}{2} H_2O_{(l)}$

• x = 1
• y = 4

$\tt CH_{4 \: (g)} + \bigg(1+ \dfrac{4}{4} \bigg) O_{2 \: (g)} \longrightarrow 1CO_{2\:(g) } + \dfrac{4}{2} H_2O_{(l)}$

$\tt CH_{4 \: (g)} + 2 O_{2 \: (g)} \longrightarrow CO_{2\:(g) } + 2 H_2O_{(l)}$

Moles of methane:

$\implies n_{(CH_4)} = \dfrac{4}{16}$

$\implies n_{(CH_4)} = \dfrac{1}{4} \: mol$

1 mole of methane produce 2 moles of water

Therefore, 1/4 moles of methane will produce moles of water = 2 × 1/4 = 1/2 mol

1 mole of water has mass = 18 g

Therefore, 1/2 mole of water has mass = 18 × 1/2 = 9 g