According to the following reaction, how many grams of carbon tetrachloride are required for the complete reaction of 21.2 grams of methane (CH4)? methane (CH4) (g) + carbon tetrachloride (g) dichloromethane (CH2Cl2) (g)

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"## 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 , the answer will be

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