Question & Answer

Determining the Limiting Reagent and Amount of H2O Produced

Given the balanced chemical equation:

$2{\text{H}}_2+{\text{O}}_2\to \mathrm{<br>}\text{<span class="base" style="font-family: "Helvetica Neue", Helvetica, Arial, sans-serif;"><span class="mord">2</span><span class="mord"><span class="mord text"><span class="mord">H</span></span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist"><span class="pstrut"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist"></span></span></span></span></span><span class="mord text"><span class="mord">O</span></span></span>}$

You start with:

• 4 moles of $\text{H}_2$
• 3 moles of $\text{O}_2$

Determine the Limiting Reagent:

According to the balanced equation:

• • 2 moles of $\text{H}_2$react with 1 mole of $\text{O}_2$.

To determine the limiting reagent, calculate how much $\text{O}_2$is needed for 4 moles of $\text{H}_2$:

• For 4 moles of $\text{H}_2$, the required $\text{O}_2$is $\frac{4 \text{ moles H}_2}{2} = 2 \text{ moles O}_2$.

You have 3 moles of $\text{O}_2$, which is more than the 2 moles required. Therefore, $\text{H}_2$ is the limiting reagent.

Amount of $\text{H}_2\text{O}$Produced:

From the balanced equation, 2 moles of $\text{H}_2$produce 2 moles of $\text{H}_2\text{O}$.

Therefore, 4 moles of $\text{H}_2$will produce: $\frac{4 \text{ moles H}_2}{2} \times 2 \text{ moles H}_2\text{O} = 4 \text{ moles H}_2\text{O}$.

So, the limiting reagent is $\text{H}_2$, and the amount of $\text{H}_2\text{O}$ produced is 4 moles.

. Molarity Calculation

Given:

• Volume of the solution = 500 mL = 0.5 L
• Concentration of $\text{NaCl= 0.2M}$
• $\text{<br>}$

• $\text{moles} = \text{Molarity} \times \text{Volume (in liters)}$

$\text{moles NaCl}=0.2\text{ M}\times 0.5\text{ L}=0.1\text{ moles NaCl}$

Mass of $\text{<strong style="font-family: "Helvetica Neue", Helvetica, Arial, sans-serif;"><span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="mord text"><span class="mord">NaCl</span></span></span></span></span> Required:</strong>}$

The molar mass of $\text{NaCl}$ = 58.44g/mol

The mass can be calculated using: $\text{mass}=\text{moles}\times \text{molar mass}$

$\text{mass NaCl} = 0.1 \text{ moles} \times 58.44 \text{ g/mol} = 5.844 \text{ grams}$

Summary