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How many milligrams of magnesium reacts with excess HCl to produce 31.2 mL of hydrogen gas at 754 Torr and 25.0℃.

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Answer:

There will react 30.9 milligrams of magnesium

Step-by-step explanation:

Step 1: Data given

Volume of hydrogen = 31.2 mL

Pressure = 754 torr = 754/760 = 0.992 atm

Temperature = 25.0 °C = 298 Kelvin

Step 2: The balanced equation

Mg + 2HCl → MgCl2 + H2

Step 3: Calculate moles of H2

p*V = n*R*T

⇒ with p = the pressure of H2 = 0.992 atm

⇒with V = the volume of H2 = 31.2 mL = 0.0312 L

⇒ with n = the moles of H2 = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T= the temperature = 25.0 °C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.992*0.0312)/(0.08206*298)

n = 0.00127 moles

Step 4: Calculate moles of Mg

For 1 mol of Mg we need 2 moles of HCl to produce 1 mol of MgCl2 and 1 mol of H2

For 0.00127 moles of H2 we need 0.00127 moles of Mg

Step 5: Calculate mass of Mg

Mass of Mg = moles of Mg * molar mass of Mg

Mass of Mg = 0.00127 moles * 24.3 g/mol

Mass of Mg = 0.0309 grams = 30.9 mg of Mg

There will react 30.9 milligrams of magnesium

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