Question

When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of 3.8 M HCl is required to react completely with 3.33 g of magnesium?

A) 0.025 L
B) 0.05 L
C) 0.1 L
D) 0.2 L

Answer 1

A because that we can be hydrated if we just drink 0.2L hope it helpss

Answer 2

To find the volume of 3.8 M HCl needed to react with 3.33 g of magnesium, we calculate the moles of magnesium, use the mole ratio from the balanced equation to find the moles of HCl needed, and then use the molarity to find the volume. Approximately 50 mL of HCl is required, corresponding with option B) 0.05 L.

Step-by-step explanation:

The question concerns a chemical reaction between hydrochloric acid (HCl) and magnesium (Mg), which produces hydrogen gas (H₂) and magnesium chloride (MgCl₂) in aqueous solution. To determine the volume of a 3.8 M HCl solution needed to react with 3.33 g of magnesium, we need to follow a series of stoichiometric steps:

1. Calculate the moles of magnesium by using the molar mass of magnesium (24.305 g/mol).
2. Use the balanced chemical equation, Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g), to find the mole ratio between magnesium and hydrochloric acid.
3. Calculate the moles of HCl needed using the mole ratio.
4. Calculate the volume of HCl solution using its concentration (Molarity).

Step 1: Moles of Mg = 3.33 g / 24.305 g/mol = 0.1369 mol Mg

Step 2: The mole ratio of Mg to HCl from the balanced equation is 1:2.

Step 3: Moles of HCl = 0.1369 mol Mg * 2 mol HCl/mol Mg = 0.2738 mol HCl

Step 4: Volume of HCl = Moles of HCl / Molarity of HCl = 0.2738 mol HCl / 3.8 M = 0.07205 L

This volume is closest to option B) 0.05 L, therefore, approximately 50 mL of 3.8 M HCl is required.