Question

use the references to access important values if needed for this question.
what volume of carbon tetrachloride is required to react completely with 2.42 mol of methane (ch4) according to the following reaction at 0°c and 1 atm?
methane (ch4)(g) + carbon tetrachloride(g) → dichloromethane (ch2cl2)(g)
volume = 1 l
retry entire group
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Explanation:

Step1: Determine the mole ratio

From the reaction \(CH_{4}(g)+CCl_{4}(g)\to2CH_{2}Cl_{2}(g)\), the mole ratio of \(CH_{4}\) to \(CCl_{4}\) is \(1:1\). So, if there are \(2.42\) mol of \(CH_{4}\), we need \(2.42\) mol of \(CCl_{4}\).

Step2: Use the ideal gas law \(PV = nRT\)

We know \(P = 1\) atm, \(n=2.42\) mol, \(R = 0.0821\space L\cdot atm/(mol\cdot K)\), and \(T=(0 + 273)K=273\) K. Rearranging the ideal gas law for \(V\) gives \(V=\frac{nRT}{P}\).
Substitute the values: \(V=\frac{2.42\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times273\space K}{1\space atm}\)

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Answer:

\(54.2\space L\)