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Chapter 4: Problem 116

When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of 5.0 M HCl is required to react completely with 3.00 g of magnesium?

Short Answer

Expert verified
To react completely with 3.00 g of magnesium, 49.2 mL of 5.0 M HCl is required.

Step by step solution

01

1. Write the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between magnesium (Mg) and hydrochloric acid (HCl) is given by: \[Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)\]
02

2. Convert the mass of magnesium to moles.

To convert the mass of magnesium to moles, we will use the molar mass of magnesium (24.31 g/mol). The formula to calculate the moles of magnesium is: Moles of Mg = (mass of Mg) / (molar mass of Mg) Moles of Mg = \(\frac{3.00\, g}{24.31\, g/mol} = 0.123\, mol\)
03

3. Use stoichiometry to find the moles of HCl required for the reaction.

From the balanced chemical equation, we observe that the stoichiometric ratio of Mg to HCl is 1:2. This means that for every one mole of Mg, two moles of HCl are required. Therefore, we can calculate the moles of HCl needed, using the moles of Mg: Moles of HCl = 2 × (moles of Mg) Moles of HCl = 2 × (0.123 mol) = 0.246 mol
04

4. Calculate the volume of 5.0 M HCl solution needed.

We are given the concentration of HCl solution, which is 5.0 M (moles per liter). We can use this information to calculate the volume of HCl solution required by using the formula: Volume of HCl solution = (moles of HCl) / (concentration of HCl solution) Volume of HCl solution = \(\frac{0.246\, mol}{5.0\, M} = 0.0492\, L\) To get the volume in milliliters, we can convert liters to milliliters: Volume of HCl solution = 0.0492 L × 1000 mL/L = 49.2 mL So, 49.2 mL of 5.0 M HCl is required to react completely with 3.00 g of magnesium.

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