Question

What volume of hydrogen gas, in mL, is produced from 0.128 g Mg reacting with an excess of HCI at a pressure of 1.28 atm and a temperature of 312 K?

Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)​

Answer 1

To determine the volume of hydrogen gas produced in the reaction, we need to use the stoichiometry of the balanced equation and the ideal gas law. The volume of hydrogen gas produced from 0.128 g of Mg reacting with an excess of HCl at a pressure of 1.28 atm and a temperature of 312 K is approximately 127 mL.

Step-by-step explanation:

To determine the volume of hydrogen gas produced in the reaction, we need to use the stoichiometry of the balanced equation. The balanced equation shows that 1 mole of magnesium (Mg) reacts with 2 moles of hydrochloric acid (HCl) to produce 1 mole of hydrogen gas (H₂). First, we need to convert the mass of Mg to moles using its molar mass, which is 24.305 g/mol. Next, we can use the moles of Mg to calculate the moles of H₂ and then convert to volume using the ideal gas law.

Given: Mass of Mg = 0.128 g, Pressure = 1.28 atm, Temperature = 312 K.

Step 1:

Convert mass of Mg to moles:

0.128 g Mg × (1 mol Mg / 24.305 g Mg) = 0.00527 mol Mg

Step 2:

Use stoichiometry to find moles of H₂:

0.00527 mol Mg × (1 mol H₂ / 1 mol Mg) = 0.00527 mol H₂

Step 3:

Use the ideal gas law to find the volume of H₂:

V = (nRT) / P = (0.00527 mol)(0.0821 L·atm/mol·K)(312 K) / 1.28 atm ≈ 0.127 L = 127 mL

Answer 2

132 mL of hydrogen gas is produced.

Step-by-step explanation:

0.128 g Mg * (1 mol Mg/24.31 g Mg) * (1 mol H2/1 mol Mg) = 0.00526 mol H2

P = 1.28 atm

V = ?

n = 0.00526 mol

R = 0.08206 L·atm/mol·K

T = 312 KV = nRT/P

V = (0.00526 mol)(0.08206 L·atm/mol·K)(312 K) / (1.28 atm)

V = 0.132 L = 132 mL