Question

what mass of H2SO4 is required to react with 60.0mL of 0.233 M Al(OH)3 solution? (H2SO4= 98.09 g/mol) balanced equation:2Al(OH)3 + 3H2SO4 —> Al2(SO4)3 + 6H2O

Answer 1

The mass of H2SO4 is 2.06 grams

Step-by-step explanation

Given that;

The volume of Al(OH)3 is 60.0mL

The molarity of the solution is 0.233M

Follow the steps below to find the mass of H2SO4

Step 1; Write the balanced equation of the reaction

`\text{ 2Al\lparen OH\rparen}_3\text{ + 3H}_2SO_4\text{ }\rightarrow\text{ Al}_2(SO_4)_3\text{ + 6H}_2O`

Step 2; Find the number of moles of Al(OH)3 using the below formula

`\text{ molarity = }\frac{\text{ no of moles }}{\text{ volume of solution}}`

Convert the volume of the solution to liters

`\begin{gathered} \text{ 1mL }\rightarrow\text{ 0.001L} \\ \text{ 60mL }\rightarrow\text{ xL} \\ \text{ cross multiply} \\ \text{ 1mL }*\text{ xL = 60mL }*\text{ 0.001L} \\ \text{ Isolate x} \\ xL\text{ = }\frac{60\cancel{mL}*\text{ 0.001L}}{1\cancel{mL}} \\ \text{ xL = 60 }*\text{ 0.001L} \\ \text{ x = 0.06L} \end{gathered}`

The volume of the solution in Liters is 0.06L

Substitute the given data into the formula in step 2

`\begin{gathered} \text{ 0.233 = }\frac{\text{ moles of solute}}{\text{ 0.06}} \\ \text{ cross multiply} \\ \text{ moles of solute = 0.233}*\text{ 0.06} \\ \text{ moles of solute = 0.01398 mol} \end{gathered}`

Therefore, the number of moles of Al(OH)3 is 0.01398 mol

Step 3; Find the number of moles of H2SO4 using a stoichiometry ratio

Let x represents the number of moles of H2SO4

`\begin{gathered} \text{ 2 moles Al\lparen OH\rparen}_3\text{ }\rightarrow\text{ 3 moles H}_2SO_4 \\ \text{ 0.01398 mol Al\lparen OH\rparen}_3\text{ }\rightarrow\text{ x moles H}_2SO_4 \\ \text{ Cross multiply} \\ \text{ 2 moles Al\lparen OH\rparen}_3\text{ }*\text{ x moles H}_2SO_4\text{ = 0.01398 mol Al\lparen OH\rparen}_3*\text{ 3 moles H}_2SO_4 \\ \text{ Isolate x} \\ x\text{ = }\frac{0.01398mol\cancel{Al(OH)\placeholder{⬚}_3}*\text{ 3 moles H}_2SO_4}{2moles\cancel{Al(OH)\placeholder{⬚}_3}} \\ \text{ x = }\frac{0.01398*\text{ 3}}{2} \\ \text{ x = 0.02097 mol} \end{gathered}`

Therefore, the number of moles of H2SO4 is 0.02097 mol

Step 4; Find the mass of H2SO4 using the below formula

`\begin{gathered} \text{ mole = }\frac{\text{ mass}}{\text{ molar mass}} \\ \text{ cross multiply} \\ \text{ mass = mole }*\text{ molar mass} \end{gathered}`

Recall, that the molar mass of H2SO4 is 98.09 g/mol

`\begin{gathered} \text{ mass = 0.02097 }*\text{ 98.09} \\ \text{ mass = 2.06 grams} \end{gathered}`

Hence, the mass of H2SO4 is 2.06 grams