For the equation below: Mg(s)+2HCl(aq)--->MgCl₂(aq)+H₂(g). How many moles of hydrogen will be liberated from 177.6mL of 4.94MHCl reacting with an excess of magnesium?

To determine how many moles of hydrogen gas \((\(\text{H}_2\)) will be liberated, we first need to calculate the moles of hydrochloric acid (\(\text{HCl}\)) that are reacting. We use the molarity (M) and volume (V) of the \(\text{HCl}\) solution.\)

Step-by-Step Calculation

1. Calculate the moles of HCl:

\(\[\text{Moles of HCl} = \text{Molarity (M)} \times \text{Volume (L)}\]\)

Given:

\(\[ \text{Molarity} = 4.94\, \text{M} \] \[ \text{Volume} = 177.6\, \text{mL} = 0.1776\, \text{L} \]\)

Plugging in the values:

\(\[\text{Moles of HCl} = 4.94\, \text{M} \times 0.1776\, \text{L} = 0.877344\, \text{moles}\]\)

2. Determine the moles of \(\(\text{H}_2\)\) produced:

The balanced chemical equation is:

Mg(s) + 2HCl(aq) \(\rightarrow\) \(MgCl_2(aq)+\{H}_2(g)\]\)

From the equation, we see that 2 moles of HCl produce 1 mole of \(\(\text{H}_2\). Therefore, the moles of \(\text{H}_2\) produced can be calculated using the stoichiometry of the reaction:\)

\(\[\text{Moles of } \text{H}_2 = \frac{\text{Moles of HCl}}{2}\]\)

Plugging in the moles of \(\text{HCl}\):

\(\[\text{Moles of } \text{H}_2 = \frac{0.877344}{2} = 0.438672\]\)

Answer

The number of moles of hydrogen gas \((\(\text{H}_2\)) liberated is:\[0.438672 \text{ moles}\]\)

sriLEKHA89 sriLEKHA89 · Answer 2 · 2024-06-04T12:47:24-04:00

The moles of hydrogen liberated is 0.438672 mol.

First, we need to calculate the moles of HCl. Using the formula:

Moles of HCl = Molarity (M) × Volume (L)

Convert 177.6 mL to liters:

177.6 mL × (1 L / 1000 mL) = 0.1776 L

Now, calculate the moles of HCl:

Moles of HCl = 4.94 M × 0.1776 L = 0.877344 moles

From the balanced equation, 2 moles of HCl produce 1 mole of H₂.

So, the moles of H₂ produced will be:

Moles of H₂ = 0.877344 moles HCl × (1 mole H₂ / 2 moles HCl) = 0.438672 moles H₂