Question

Question #9f667

Answer 1

Your reaction needs `"203 g"` of carbon tetrachloride.

Start with the balanced chemical equation

`CH_(4(g)) + "CCl"_(4(l)) -> 2CH_2Cl_(2(g))`

Notice that you have a `"1:1"` mole ratio between methane and carbon tetrachloride, which means that, for every 1 mole of the former, you'll need exactly 1 mole of the latter.

You can determine how many moles of methane you have by using its molar mass

`"21.2 g CH"_4 * "1 mole"/"16.04 g" = "1.322 moles CH"_4`

This means that the number of moles of carbon tetrachloride needed is

`"1.322 moles CH"_4 * "1 mole CCl"_4/"1 mole CH"_4 = "1.322 moles CCl"_4`

Now use carbon tetrachloride's molar mass to calculate how many grams are needed

`"1.322 moles CCl"_4 * "153.82 g"/"1 mole" = "203.35 g"` `"CCl"_4`

Rounded to three sig figs, the answer will be

`m_("CCl"_4) = "203 g"`