Question

Calculate the mass (in grams) of magnesium metal needed to produce 43.0 ml of hydrogen gas at an atmospheric pressure of 739 mmhg and a temperature of 25.9°c.

Answer 1

To calculate the required mass of magnesium for producing 43.0 ml of hydrogen gas at 739 mmHg and 25.9°C, we use the ideal gas law to find moles of hydrogen and equate it to moles of magnesium. This calculation gives us 0.109 grams of magnesium needed.

Step-by-step explanation:

To produce 43.0 ml of hydrogen gas at 739 mmHg and 25.9°C, you will need to use 0.109 grams of magnesium metal.

To calculate the mass of magnesium needed, we first use the ideal gas law (PV=nRT) to find the number of moles of hydrogen (H₂)

Convert volume to liters: 43.0 ml = 0.043 L

Convert temperature to Kelvin: T = 25.9 + 273.15 = 298.05 K

Convert pressure to atmospheres: P = 739 mmHg * (1 atm / 760 mmHg) = 0.97263 atm

Universal Gas Constant R = 0.08206 L·atm/K·mol

Using the ideal gas law: n = PV / RT = (0.97263 atm * 0.043 L) / (0.08206 L·atm/K·mol * 298.05 K) = 0.0018 mol

From the chemical equation: 1 mol Mg = 1 mol H₂. Therefore, moles of Mg and moles of H₂ are equal.

Using molar masses: 0.0018 mol * 24.305 g/mol = 0.109 g of Mg