Answer:
1) 1.235 g.
2) 0.61 g.
Explanation:
- From the balanced equation:
<em>Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O.</em>
1.0 mol of Al(OH)₃ reacts with 3.0 moles of HCl to produce 1.0 mol of AlCl₃ and 3.0 moles of H₂O.
<em>1) How many grams of HCl can a tablet with 0.880 g of Al(OH)₃ consume? </em>
- To calculate the amount of HCl needed to consume 0.880 g of Al(OH)₃, we need to calculate the no. of moles of Al(OH)₃:
no. of moles of Al(OH)₃ = mass/molar mass = (0.880 g)/(78.0 g/mol) = 1.13 x 10⁻² mol.
∵ Every 1.0 mol of Al(OH)₃ needs 3.0 moles of HCl to be consumed.
∴ 1.13 x 10⁻² mol of Al(OH)₃ needs (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of HCl.
The no. of grams of HCl = no. of moles of HCl x molar mass of HCl = (3.385 x 10⁻² mol)(36.5 g/mol) = 1.235 g.
<em>2) How much H₂O?</em>
∵ Every 1.0 mol of Al(OH)₃ produces 3.0 moles of H₂O.
∴ 1.13 x 10⁻² mol of Al(OH)₃ produces (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of H₂O.
<em>The no. of grams of H₂O = no. of moles of H₂O x molar mass of H₂O </em>= (3.385 x 10⁻² mol)(18.0 g/mol) = <em>0.6092 g ≅ 0.61 g.</em>
