Question

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

Answer 1

The mass of AlCl3 is 1250.86 grams

The mass of LiH is 298.30785 grams

Step-by-step explanation

Given reaction

`\text{ 4LiH + AlCl}_3\text{ }\rightarrow\text{ LiAlH}_4+\text{ 3LiCl}`

The mass of LiAlH4 is 356 grams

To find the mass of the reactants, follow the steps below

Step 1: Find the number of moles of LiAlH4 using the below formula

`\text{ Mole = }\frac{\text{ mass}}{\text{ molar mass}}`

Recall, that the molar mass of LiAlH4 is 37.95 g/mol as provided in the periodic table

`\begin{gathered} \text{ Mole = }\frac{\text{ mass}}{\text{ molar mass}} \\ \text{ Mole = }(356)/(37.95) \\ \text{ Mole = 9.381 moles} \end{gathered}`

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

`\begin{gathered} 4\text{ }\rightarrow\text{ 1} \\ x\text{ }\rightarrow\text{ 9.381} \\ \text{ cross multiply} \\ x*1\text{ = 4 }*\text{ 9.381} \\ x\text{ = 37.523 moles} \end{gathered}`

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

`\begin{gathered} 1\text{ }\rightarrow\text{ 1} \\ x\text{ }\rightarrow9.381 \\ cross\text{ mulitiply} \\ 1\text{ }*\text{ 9.381 = x }*1 \\ 9.381\text{ = x} \\ x\text{ = 9.381 moles} \end{gathered}`

The number of moles of AlCl3 is 9.381 moles

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

`\text{ Mole= }\frac{\text{ mass}}{\text{ molar mass}}`

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

`\begin{gathered} \text{ Mole = }\frac{mass}{molar\text{ mass}} \\ \text{ mole = 37.523 moles, and molar mass = 7.95 g/mol} \\ \text{ 37.523 = }(mass)/(7.95) \\ mass\text{ = 37.523 }*\text{ 7.95} \\ \text{ mass = 298.30785 grams} \end{gathered}`

Hence, the mass of LiH is 298.30785 grams

For AlCl3

`\begin{gathered} \text{ Mole = }\frac{mass}{molar\text{ mass}} \\ \text{ mole = 9.381 moles, molar mass = 133.34 g/mol} \\ \text{ 9.381 = }(mass)/(133.34) \\ \text{ cross multiply} \\ \text{ Mass = 9.381}*133.34 \\ \text{ Mass = 1250.86 grams} \end{gathered}`

The mass of AlCl3 is 1250.86 grams