Question

Based on the balanced equation:3 Ag2S + 2 Al -> Al2S3 + 6 Ag143.2 grams of Ag2S = ? Moles of Al2S3

Answer 1

ANSWER

the number of moles of Al2S3 is 0.193 mol

EXPLANATION

Given that:

The mass o Ag2S in grams = 143.2 grams

Write the balanced equation for the reaction

$\text{ 3Ag}_2S\text{ + 2Al }\rightarrow\text{ Al}_2S_3\text{ + 6Ag}$

Find the number of moles of Ag2S using the below formula

$\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}$

Recall, that the molar mass of Ag2S 247.8 g/mol

$\begin{gathered} \text{ mole = }\frac{\text{ 143.2}}{247.8} \\ \\ \text{ mole = 0.578 mol} \end{gathered}$

The mole of Ag2S is 0.578 mol

Find the mole of Al2S3 using a stoichiometry ratio

In the balanced equation, 3 moles of Ag2S give 1 moles Al2S3

$\begin{gathered} \text{ 3 moles Ag}_2S\text{ }\rightarrow\text{ 1 mole Al}_2S_3 \\ \text{ 0.578 mol }\rightarrow\text{ x mole Al}_2S_3 \\ \text{ cross multiply} \\ \text{ 3 moles }\times\text{ x moles Al}_2S_3\text{ }=\text{ 1 mole Al}_2S_3\times\text{ 0.578} \\ \text{ isolate x} \\ \text{ x = }\frac{1\text{ }\times\text{ 0.578}}{3} \\ \text{ x = 0.193 mol} \end{gathered}$

Therefore, the number of moles of Al2S3 is 0.193 mol