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The ccl4 formed in the first step is used as a reactant in the second step. if 2.00 mol of ch4 reacts, what is the total amount of hcl produced? assume that cl2 and hf are present in excess.

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Kalahira
CH4 + 4 Cl2 → CCl4 + 4 HCl (4.00 mol CH4) x (1/1) x (0.70) = 2.80 mol CCl4 (4.00 mol CH4) x (4/1) x (0.70) = 11.2 mol HCl CCl4 + 2 HF → CCl2F2 + 2 HCl (2.80 mol CCl4) x (2/1) x (0.70) = 3.92 mol HCl 11.2 mol + 3.92 mol = 15.1 mol HCl from both steps

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