How many grams of carbon dioxide are produced from the combustion of 1.3 moles of acetylene

1 answer:

8 0

Complete combustion of acetylene generated carbon dioxide and water. This can be represent by following reaction

2C2H2 + 5O2 → 4CO2 + 2H2O
(2 mole) (4 mole)

From the above balanced reaction, it can be seen that 2 mole of acetylene on complete combustion generates 4 moles of carbon dioxide
i.e. 2 mole of C2H2 ≡ 4 mole of CO2
∴ 1.3 mole of C2H2 ≡ (4 X 1.3)/2 = 2.6 mole of CO2

Now, 1 mole of CO2 = 44 g
∴ 2.6 mole of CO2 = (44 X 2.6) = 114.4 g

Thus, <span>114.4 grams of carbon dioxide are produced from the combustion of 1.3 moles of acetylene</span>

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Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

$E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}$

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$E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\$