Question

9. How many moles of H2SO4 are required to completely react with 8.3 mol Al?

Answer 1

12.45 moles of H2SO4 will reacts completely with 8.3 mol of Al

Step-by-step explanation

Given information

The number of moles of Al is 8.3 mol

To find the number of moles of H2SO4 that reacts with Al, you will need to follow the steps below

Step 1: Write a balanced equation between the reaction Al and H2SO4

`\text{ 2Al}_((s))\text{ + 3H}_2SO_(4(aq))\rightarrow Al_2(SO_4)_(3(aq))\text{ + 3H}_(2(g))`

The aim of the above reaction is to produce hydrogen gas. From the above reaction, you will see that 2 moles of aluminum react with 2 moles of H2SO4.

Since we already know the number of moles of Al, therefore, we can calculate the number of moles of H2SO4 using a stoichiometric ratio

Step 2: Find the number of moles of H2SO4

Let x represents the number of moles of H2SO4

According to their stoichiometric ratio; 2 moles of Al react with 3 moles of H2SO4

`\begin{gathered} \text{ 2 moles of Al }\rightarrow\text{ 3 moles of H}_2SO_4 \\ \text{ 8.3 moles of Al }\rightarrow\text{ x moles of H}_2SO_4 \\ \text{ cross multiply} \\ \text{ 2 moles of Al }*\text{ x moles of H}_2SO_4\text{ = 3 moles of H}_2SO_4*\text{ 8.3 moles of Al} \\ \text{ x moles of H}_2SO_4\text{ = }\frac{3\text{ moles of H}_2SO_{4\text{ }}*8.3\cancel{moles\text{ of Al}}}{2\cancel{moles\text{ of Al}}} \\ \text{ x moles of H}_2SO_4\text{ = }\frac{3\text{ }*8.3}{2}\text{ moles of H}_2SO_(_4) \\ \text{ x moles of H}_2SO_4\text{ = }(24.9)/(2)\text{ moles of H}_2SO_4 \\ \text{ x moles of H}_2SO_4\text{ = 12.45 moles} \end{gathered}`

Hence, 12.45 moles of H2SO4 will reacts completely with 8.3 mol of Al