Question

According to the following reaction, how many moles of dichloromethane (CH2Cl2) will be formed upon the complete reaction of 0.766 moles methane (CH4) with excess carbon tetrachloride?

methane (CH4) (g) + carbon tetrachloride (g) → dichloromethane (CH2Cl2) (g)

Answer 1

Answer:

$1.53~moles~CH_2Cl_2$.

Explanation:

We can start with the reaction, if we know the formula for each compound:

-) Methane: $CH_4$

-) Carbon tetrachloride: $CCl_4$

-) Dichloromethane: $CH_2Cl_2$

With this in mind, we can write the reaction:

$CH_4~+~CCl_4~->~CH_2Cl_2$

Now, we can balance the reaction:

$CH_4~+~CCl_4~->~2CH_2Cl_2$

After this, we have 2 carbon atoms on each side, 4 hydrogen atom on each side, and 4 chlorine atoms on each side.

If we want to know how many moles of $CH_2Cl_2$ would be produced with .766 moles of $CH_4$, we have to check the balanced reaction and use the molar ratio. In this case, the molar ratio is 1 mol $CH_4$ will produce 2 moles of $CH_2Cl_2$ (1:2). So:

$0.766~moles~CH_4\frac{2~moles~CH_2Cl_2}{1~mol~CH_4}=1.53~moles~CH_2Cl_2$

We wil have $1.53~moles~CH_2Cl_2$.

I hope it helps!