Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. A \(10.00-\mathrm{g}\) mixture of zinc and magnesium produces \(0.5171 \mathrm{~g}\) of hydrogen gas upon being mixed with an excess of hydrochloric acid. Determine the percent magnesium by mass in the original mixture.

For the reaction of zinc and hydrochloric acid, we have: \( Zn + 2HCl \rightarrow ZnCl_2 + H_2 \) For the reaction of magnesium and hydrochloric acid, we have: \( Mg + 2HCl \rightarrow MgCl_2 + H_2 \)

From the balanced chemical equations, we can see that 1 mole of zinc reacts to produce 1 mole of hydrogen gas, and the same applies to magnesium. We'll first find out how many moles of hydrogen gas were produced. To do this, we'll use the molar mass of hydrogen gas which is approximately 2 g/mol: Moles of \(H_2\) produced = Mass of \(H_2\) / Molar mass of \(H_2\) = \(0.5171 \mathrm{~g} / 2 \mathrm{~g/mol}\) = 0.25855 moles. Since the moles of hydrogen gas produced are equal to the moles of zinc and magnesium involved in the reaction, we have: \( moles(Zn) + moles(Mg) = 0.25855 \) Now, let x be the mass of magnesium in the mixture, so the mass of zinc will be (10 - x). We can now find the moles of each metal: \( moles(Zn) = \frac{(10 - x)}{65.38} \) (We use 65.38 g/mol as the molar mass of zinc) \( moles(Mg) = \frac{x}{24.31} \) (We use 24.31 g/mol as the molar mass of magnesium)

We can now write the following equation, based on the moles of zinc and magnesium: \(\frac{(10 - x)}{65.38} + \frac{x}{24.31} = 0.25855 \) Now, we will solve the above equation for x to obtain the mass of magnesium in the original mixture: \(x = 2.0775 \mathrm{~g}\)

We have now found the mass of magnesium in the original mixture. To determine the mass percent of magnesium, we'll divide the mass of magnesium by the total mass and multiply by 100: Percent magnesium by mass = \(\frac{2.0775 \mathrm{~g}}{10 \mathrm{~g}} \times 100\% = 20.775 \% \) Therefore, the percent magnesium by mass in the original mixture is approximately 20.77%.