Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# 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.

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"