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Magnesium metal (0.100 mol) and hydrochloric acid (0.500 mol HCl) are combined and react to completion. What volume of hydrogen gas, measured at STP, is produced? Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) (R = 0.08206 L • atm/K • mol) Select one: A. 22.4 L of H2 B. 5.60 L of H2 C. 4.48 L of H2 D. 11.2 L of H2 E. 2.24 L of H2

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Answer:

A. [tex]V=2.24L[/tex] of [tex]H_{2}[/tex]

Explanation:

1. First take the balanced chemical equation to obtain hydrogen gas:

[tex]Mg(s)+2HCl(aq)=MgCl_{2}(aq)+H_{2}(g)[/tex]

2. Find the limiting reagent to how much hydrogen is formed:

Take the number of moles of each compound and divide it between the stoichiometric coefficient in the balanced chemical equation.

- For the Mg:

[tex]\frac{0.100}{1}[/tex]

- For the HCl:

[tex]\frac{0.500}{2}=0.25[/tex]

The magnesium is the limiting reagent.

3. Find the number of moles of hydrogen produced:

[tex]0.100molesMg\frac{1molH_{2}}{1molMg}=0.100molesH_{2}[/tex]

4. Find the volume of hydrogen using the ideal gas equation at STP:

The ideal gas equation is expressed as [tex]PV=nRT[/tex], where P is the pressure, at STP P=1 atm, V is the volume, n is the number of moles, R is a constante whose value is R=0.08206[tex]\frac{L.atm}{K.mol}[/tex], and T is the temperature, at STP T=273K

Solving for V:

[tex]V=\frac{nRT}{P}[/tex]

Replacing values:

[tex]V=\frac{0.100moles*0.08206\frac{L.atm}{K.mol}*273K}{1atm}[/tex]

[tex]V=2.24L[/tex] of [tex]H_{2}[/tex]

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