Question

Determine the mass in grams of HCl that can react with 0.750 g of Al(OH)$_3$ according to the following reaction Al(OH)$_3$(s) + 3 HCl(aq) ? AlCl$_3$(aq) + 3 H$_2$O(aq)

          Determine the mass in grams of HCl that can react with 0.750 g of Al(OH)$_3$ according to the following reaction
Al(OH)$_3$(s) + 3 HCl(aq) ? AlCl$_3$(aq) + 3 H$_2$O(aq)

Added by Vanessa G.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Ronald Prasad

10/08/2022

Step 1

0.750 g Al(OH)3 × (1 mol Al(OH)3 / 78.00 g Al(OH)3) = 0.00961 mol Al(OH)3 Show more…

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What should I fill in the blank? Determine the mass in grams of HCl that can react with 0.750 g of Al(OH)3 according to the following reaction: Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(aq) Starting Amount Al(OH)3: 0.750 g Molar Mass Al(OH)3: 78.00 g/mol HCl: 36.46 g/mol H2O: 18.02 g/mol Molar Ratios 1 mol Al(OH)3 : 3 mol HCl 1 mol HCl : 3 mol H2O Calculations: Step 1: Convert grams of Al(OH)3 to moles 0.750 g Al(OH)3 × (1 mol Al(OH)3 / 78.00 g Al(OH)3) = 0.00961 mol Al(OH)3 Step 2: Use the mole ratio to find moles of HCl 0.00961 mol Al(OH)3 × (3 mol HCl / 1 mol Al(OH)3) = 0.0288 mol HCl Step 3: Convert moles of HCl to grams 0.0288 mol HCl × (36.46 g HCl / 1 mol HCl) = 1.05 g HCl