madey21
Answered

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[tex]\[
AgNO_3 + NaCl \rightarrow AgCl + NaNO_3
\][/tex]

75.0 mL of 0.500 M \(AgNO_3\) reacts with 35.0 mL of 1.00 M \(NaCl\). How many moles of \(AgCl\) can form from the \(NaCl\)?

[tex]\[
\frac{0.0350 \, \text{L} \, NaCl}{1} \times \frac{1.00 \, \text{mol} \, NaCl}{1 \, \text{L}} \times \frac{1 \, \text{mol} \, AgCl}{1 \, \text{mol} \, NaCl} =
\][/tex]
[tex]\[
[?] \, \text{mol} \, AgCl
\][/tex]

Sagot :

GinnyAnswer
Let's solve this problem step-by-step to determine how many moles of \(AgCl\) can be formed from the given amount of \(NaCl\):

1. Identify the key information:
- Volume of \(NaCl\) solution \((V) = 35.0\) mL
- Molarity of \(NaCl\) solution \((M) = 1.00\) M

2. Convert the volume of \(NaCl\) solution from milliliters to liters:
[tex]\[ V_{NaCl} = 35.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.0350 \, \text{L} \][/tex]

3. Calculate the moles of \(NaCl\) using its molarity and volume:
[tex]\[ \text{Moles of NaCl} = M \times V_{NaCl} \][/tex]
Plugging in the values:
[tex]\[ \text{Moles of NaCl} = 1.00 \, \text{M} \times 0.0350 \, \text{L} = 0.035 \, \text{moles} \][/tex]

4. Determine the moles of \(AgCl\) that can form:
The reaction between \(AgNO_3\) and \(NaCl\) is given by:
[tex]\[ AgNO_3 + NaCl \rightarrow AgCl + NaNO_3 \][/tex]
The stoichiometry of the reaction shows a 1:1 molar ratio between \(NaCl\) and \(AgCl\). Thus, the moles of \(NaCl\) will be directly converted to moles of \(AgCl\).

5. Moles of \(AgCl\) that can form:
[tex]\[ \text{Moles of AgCl} = \text{Moles of NaCl} = 0.035 \, \text{moles} \][/tex]

So, the number of moles of [tex]\(AgCl\)[/tex] that can form from the [tex]\(NaCl\)[/tex] is [tex]\(\boxed{0.035}\)[/tex] moles.
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