$
\begin{array}{l}
AgNO _3 + NaCl \rightarrow AgCl + NaNO _3 \\
\Delta H = -67 \, \text{kJ/mol}
\end{array}
$

Considering the limiting reactant, what is the heat change for the reaction when 0.0350 moles of [tex]$NaCl$[/tex] reacts with [tex]$AgNO _3$[/tex]?

$
q = \frac{0.0350 \, \text{mol}}{1} \times \frac{-67 \, \text{kJ}}{1 \, \text{mol}} \times \frac{1000 \, \text{J}}{1 \, \text{kJ}}
$

Question

01-07-2024
Chemistry

Answered

$
\begin{array}{l}
AgNO _3 + NaCl \rightarrow AgCl + NaNO _3 \\
\Delta H = -67 \, \text{kJ/mol}
\end{array}
$

Considering the limiting reactant, what is the heat change for the reaction when 0.0350 moles of [tex]$NaCl$[/tex] reacts with [tex]$AgNO _3$[/tex]?

$
q = \frac{0.0350 \, \text{mol}}{1} \times \frac{-67 \, \text{kJ}}{1 \, \text{mol}} \times \frac{1000 \, \text{J}}{1 \, \text{kJ}}
$

Answer :

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GinnyAnswer GinnyAnswer · Answer 1 · 2024-07-01T06:28:24-04:00

Let's solve the problem step-by-step:

1. Identify the given information:
- Moles of $ NaCl $: $ 0.0350 $ moles
- Enthalpy change ($ \Delta H $) for the reaction: $ -67 $ kJ/mol

2. Understand the relationship between the variables:
- The enthalpy change $ \Delta H $ of $ -67 $ kJ/mol indicates that each mole of $ NaCl $ reacted releases $ 67 $ kJ of energy.
- We need the heat change (q) when $ 0.0350 $ moles of $ NaCl $ react completely.

3. Set up the equation:
- Using the given formula, the heat change $ q $ can be calculated as follows:

[tex]\[ q = \frac{0.0350 \text{ mol}}{1} \times \frac{-67 \text{ kJ}}{1 \text{ mol}} \times \frac{1000 \text{ J}}{1 \text{ kJ}} \][/tex]

4. Perform the calculation:
- First, multiply moles of $ NaCl $ by the enthalpy change per mole:

[tex]\[ 0.0350 \text{ mol} \times -67 \text{ kJ/mol} = -2.345 \text{ kJ} \][/tex]

- Convert $ -2.345 $ kJ into joules by multiplying by $ 1000 $:

[tex]\[ -2.345 \text{ kJ} \times 1000 \text{ J/kJ} = -2345.0 \text{ J} \][/tex]

5. Conclusion:
- The heat change ($ q $) for the reaction when $ 0.0350 $ moles of $ NaCl $ reacts completely with $ AgNO_3 $ is $ -2345.0 $ joules.

So, the answer is that the heat change for the reaction is [tex]$ -2345.0 $[/tex] J.