. An unknown amount of magnesium metal reacts with 100.0 mL 0f 0.100M aqueous hy

Question

. An unknown amount of magnesium metal reacts with 100.0 mL 0f 0.100M aqueous hydrochloric acid. The final solution is acidic, and titrated to an endpoint using 0.150 M aqueous sodium hydroxide. The endpoint was neutralized after the addition of 16.45 mL of base. How many grams of magnesium metal were produced? The answer I found to this question was .0915 g.

In the previous problem, a gas was produced during the reaction of mixing magnesium metal with aqueous hydrochloric acid. From the data of Problem 5, how many grams of this gas were produced

Explanation / Answer

Alright, so let's work a little backwards, let's make the equations: Mg + 2HCl --> MgCl2 + H2(g) HCl + NaOH --> H2O + Na+ + Cl- Main point being that the neutralization of hydrochloric acid and sodium hydroxide is 1:1, therefore the moles of hydrochloric acid is equal to n = CV = (0.01645L)(0.150M) = 0.0024675 mol of HCl after the reaction with sodium hydroxide Next, we find the initial amount of hydrochloric acid: n = CV = (0.1L)(0.100M) = 0.01 mol of hydrochloric acid originally. Now we can subtract both amount to find the amount that reacted with magnesium: mols of HCl that reacted with magnesium = 0.01 mol - 0.0024675 mol = 0.0075325 mol of HCl that reacted with magnesium. Now we can use a little stoiciometry to find the magnesium metal: 0.0075325 mol of HCl = 1 mol of Mg/2 mol of HCl = 0.00376625 mol of Mg and we multiply this by the molar mass and put this into grams: 0.00376625 mol of Mg * 24.31 g/mol of Mg = 0.0916 grams (three significant digits), your answer was right. We use the same stoiciometric ratio to find the grams of gas, because this is the gas produced while reacting with the magnesium metal: 0.0075325 mol * 1 mol of H2 gas/ 2 mol of HCl = 0.00376625 mol of H2 * 2 g/mol of H2 gas = 0.00377 grams of H2 gas. Good luck :)

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