When 20.6 gm of methane and 39.7 gm of chlorine gas undergo a reaction that has

Question

When 20.6 gm of methane and 39.7 gm of chlorine gas undergo a reaction that has a 73.3 % yield. What mass of CH3Cl forms.? Hydrogen chloride also forms.

Explanation / Answer

1CH4 & Cl2 --> CH3Cl & HCl CH4 (16.04 g/mol) Cl2 (70.90 g/mol) CH3Cl (50.49 g/mol) HCl (36.46 g/mol) 70.90 / 16.04 = 4.4 X's as much Cl2 is needed to react with CH4,... they added an excess of Cl2,... so the limiting reagent is CH4 so using molar masses & based on 20.6 grams of CH4,.. mass of chloromethane (CH3CL) forms: 20.6 g CH4 x 50.49 g CH3Cl / 16.04 g CH4 = 64.84 grams of CH3Cl 73.30% of 64.84 g = 47.52 grams of CH3Cl mass of Hydrogen chloride forms: 20.6 g CH4 @ 36.46g HCl / 16.04 g CH4 = 46.82 grams of HCl 73.3% of 46.82 g = 34.322grams of CH3Cl

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