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What mass of CCL4 is formed by the reaction of 8g of methane w/ excess of Chlorine?

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CH4(g)+4Cl2(g)-->CCL4(g)+4HCL(g)

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  1. change 8 grams of CH4 into CCl4 using molar masses:

    8 g CH4 @ 153.82 g/mol CCl4 / 16.04 g/mol CH4 = 76.7 grams

    youir answer is 77 grams of CCl4

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