Final answer:
The volume of 0.100 M HCl that can be neutralized by 1.00 g of Mg is 0.8234 liters, which is calculated using stoichiometry based on the reaction between magnesium and hydrochloric acid.
Step-by-step explanation:
To calculate the volume of 0.100 M HCl that can be neutralized by 1.00 g of Mg, we need to use the stoichiometry of the reaction between magnesium and hydrochloric acid. The reaction is as follows:
Mg + 2HCl → MgCl₂ + H₂
First, we calculate the moles of Mg:
- Molar mass of Mg = 24.305 g/mol
- Moles of Mg = mass of Mg / molar mass of Mg
- Moles of Mg = 1.00 g / 24.305 g/mol = 0.04117 mol
From the balanced equation, 1 mole of Mg reacts with 2 moles of HCl, so:
- Moles of HCl = 2 × moles of Mg
- Moles of HCl = 2 × 0.04117 mol
- Moles of HCl = 0.08234 mol
Second, we calculate the volume of 0.100 M HCl needed:
- Volume of HCl = moles of HCl / molarity of HCl
- Volume of HCl = 0.08234 mol / 0.100 M
- Volume of HCl = 0.8234 L
So, 0.8234 liters (or 823.4 mL) of 0.100 M HCl can be neutralized by 1.00 g of Mg.