madey21
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 :

GinnyAnswer
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.

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