Question

when 30.0g of methane and 57.1g of chlorine gas undergo a reaction that has a 61.7% yield, what mass of chloromethane (CH3Cl) forms? Hydrogen chloride also forms

Answer 1

Solution :-

Balanced reaction equation

CH4 + Cl2 ------- > CH3Cl + HCl

Using the mole ratio of the both reactant lets calculate the mass of the CH3Cl that can be formed

Calculating using the mass of methane

(30.0 gCH4 * 1 mol / 16.04 g )*(1 mol CH3Cl/ 1 mol CH4)*(50.4875 g / 1 mol CH3Cl)*(61.7 % / 100%) =58.3 g CH3Cl

Now lets calculate using the given mass of Cl2

(57.1 g Cl2 * 1 mol / 70.906 g)* (1 mol CH3Cl/ 1 mol CH4)*(50.4875 g / 1 mol CH3Cl)*(61.7 % / 100%) =25.1 g CH3Cl

So the Cl2 gives the less mass of the product therefore the maximum mass of the chloromethane that can be formed is 25.1 g CH3Cl