Please look at photo mole to mole ratio of each reactant product

[tex]\begin{gathered} \text{ Mole = }\frac{\text{ mass}}{\text{ molar mass}} \\ \text{ Mole = }\frac{356}{37.95} \\ \text{ Mole = 9.381 moles} \end{gathered}[/tex]

Hence, the number of moles of LiAlH4 is 9.381 moles

Step 2: Find the number of moles of LiH and AlCl3 using a stoichiometry ratio

From the reaction, 4 moles of LiH give 1 mole of LiAlH4

Let x represents the number of moles of LiH

[tex]\begin{gathered} 4\text{ }\rightarrow\text{ 1} \\ x\text{ }\rightarrow\text{ 9.381} \\ \text{ cross multiply} \\ x\times1\text{ = 4 }\times\text{ 9.381} \\ x\text{ = 37.523 moles} \end{gathered}[/tex]

The number of moles of 37.523 moles

Find the number of moles of AlCl3?

1 mole of AlCl3 gives 1 mole LiAlH4

Let x represents the number of moles of AlCl3

[tex]\begin{gathered} 1\text{ }\rightarrow\text{ 1} \\ x\text{ }\rightarrow9.381 \\ cross\text{ mulitiply} \\ 1\text{ }\times\text{ 9.381 = x }\times1 \\ 9.381\text{ = x} \\ x\text{ = 9.381 moles} \end{gathered}[/tex]

The number of moles of AlCl3 is 9.381 moles

Step 3: Find the mass of LiH and AlCl3 using the below formula

[tex]\text{ Mole= }\frac{\text{ mass}}{\text{ molar mass}}[/tex]

Recall, that the molar mass of LiH is 7.95 g/mol, and the molar mass of AlCl3 is 133.34 g/mol as provided in the provided table.

For LiH

[tex]\begin{gathered} \text{ Mole = }\frac{mass}{molar\text{ mass}} \\ \text{ mole = 37.523 moles, and molar mass = 7.95 g/mol} \\ \text{ 37.523 = }\frac{mass}{7.95} \\ mass\text{ = 37.523 }\times\text{ 7.95} \\ \text{ mass = 298.30785 grams} \end{gathered}[/tex]

Hence, the mass of LiH is 298.30785 grams

For AlCl3

[tex]\begin{gathered} \text{ Mole = }\frac{mass}{molar\text{ mass}} \\ \text{ mole = 9.381 moles, molar mass = 133.34 g/mol} \\ \text{ 9.381 = }\frac{mass}{133.34} \\ \text{ cross multiply} \\ \text{ Mass = 9.381}\times133.34 \\ \text{ Mass = 1250.86 grams} \end{gathered}[/tex]

The mass of AlCl3 is 1250.86 grams