Question

How many grams of lithium hydride are necessary to completely react with 19.5 grams of titanium(IV) chloride in the reaction below?

Answer 1

$\begin{aligned} 4LiH + TiCl_4 \to Ti + 4LiCl + 2H_2 \end{aligned}$1 mole of TICl4​ reacts with 4 moles of LiH

1 mole of $TiCl_4$ = 48 + 4(35.5) = 190g

4 moles of $LiH$ = 4(7 + 1) = 4(8) = 32g

$\therefore$ 190g of $TiCl_4$ = 32 g of $LiH$

19.5g of $TiCl_4$ = xg

x = $\dfrac{19.5×32}{190} = 3.28g$

$\therefore$ 3.28grams of lithium hydride is necessary to completely react with 19.5 grams of titanium(IV) chloride.