Question

A student performed an experiment in the same manner as part B (magnesium+HCL). The temperature for the day was 22°C in the barometric pressure was 742MMHG. 43.5 mL of gas was collected how many grams of magnesium must've been used in the reaction?

Answer 1

0.0425 g

Step-by-step explanation:

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.

Step 1: Number of moles of the gas produced

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

Step 2: Moles of magnesium that reacted

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

Step 3: Mass of magnesium that reacted

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