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Mg + HCl → MgCl2 + H2

How many liters of hydrogen gas are produced at 298 K and 0.940 atm if 4.00 moles of hydrochloric acid react with an excess of magnesium metal?

User Howlger
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1 Answer

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Final answer:

The volume of hydrogen gas produced in the reaction is approximately 49.27 liters.

Step-by-step explanation:

Calculation of Hydrogen Gas Volume

The balanced chemical equation for the reaction is: Mg + 2HCl → MgCl2 + H2

From the equation, it is clear that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.

Given that 4.00 moles of hydrochloric acid are reacting, the number of moles of hydrogen gas produced will be 4.00/2 = 2.00 moles.

Under the given conditions of 298 K and 0.940 atm, we can use the ideal gas law to calculate the volume of hydrogen gas produced. The ideal gas law equation is: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Using the given pressure of 0.940 atm, the number of moles of hydrogen gas produced (2.00 moles), and the temperature of 298 K, we can rearrange the ideal gas law equation to solve for the volume:

V = (nRT) / P = (2.00 moles x 0.0821 L·atm/mol·K x 298 K) / 0.940 atm ≈ 49.27 liters

User Bryan Lee
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