<h3>
Answer:</h3>
0.0425 g
<h3>
Explanation:</h3>
We are given;
- Temperature, T for the day as 22°C
but, K = °C + 273.15
Thus, T = 295.15 K
- Barometric pressure, P = 742 mmHg
- Volume of the gas, V = 43.5 mL or 0.0435 L
We are required to calculate the mass of magnesium that was used in the reaction.
<h3>Step 1: Number of moles of the gas produced</h3>
Using the ideal gas equation, we can determine the number of moles
PV = nRT , where R is the ideal gas constant, 62.3637 L·mmHg/mol·K
Rearranging the formula;
n = PV ÷ RT
= (742 mmHg× 0.0435 L) ÷ (62.3637 × 295.15K)
= 0.00175 moles
<h3>Step 2: Moles of magnesium that reacted </h3>
The equation fro the reaction is;
Mg + 2HCl → MgCl₂ + H₂
Thus, 1 mole of Mg reacts to produce 1 mole of hydrogen gas
Therefore, moles of Mg = Moles of Hydrogen gas
= 0.00175 moles
<h3>Step 3: Mass of magnesium that reacted</h3>
We know that, Moles = Mass ÷ Molar mass
Thus, Mass = Moles × Molar mass
Molar mass of Mg = 24.305 g/mol
Hence;
Mass of Mg = 0.00175 moles × 24.305 g/mol
= 0.0425 g
Thus, 0.0425 g of magnesium reacted