Answer :
To determine the number of moles of [tex]\( \text{NaNO}_3 \)[/tex] produced from the complete reaction of [tex]\( 253 \, \text{g} \)[/tex] of [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex], we follow a step-by-step approach.
### Step 1: Determine the Molar Mass of [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex]
Given:
- Sodium (Na): [tex]\( 23 \, \text{g/mol} \)[/tex]
- Chromium (Cr): [tex]\( 52 \, \text{g/mol} \)[/tex]
- Oxygen (O): [tex]\( 16 \, \text{g/mol} \)[/tex]
Thus, the molar mass of [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex] is calculated as follows:
[tex]\[ \text{Molar Mass of } \text{Na}_2\text{CrO}_4 = 2 \times 23 + 52 + 4 \times 16 = 46 + 52 + 64 = 162 \, \text{g/mol} \][/tex]
### Step 2: Calculate the Number of Moles of [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex]
Now, we use the molar mass to find the moles of [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex]:
[tex]\[ \text{Moles of } \text{Na}_2\text{CrO}_4 = \frac{\text{Mass of } \text{Na}_2\text{CrO}_4}{\text{Molar Mass of } \text{Na}_2\text{CrO}_4} = \frac{253 \, \text{g}}{162 \, \text{g/mol}} = 1.5617283950617284 \, \text{mol} \][/tex]
### Step 3: Use the Reaction Stoichiometry to Find the Moles of [tex]\( \text{NaNO}_3 \)[/tex]
From the balanced chemical equation:
[tex]\[ \text{Pb}\left(\text{NO}_3\right)_2 + \text{Na}_2\text{CrO}_4 \rightarrow \text{PbCrO}_4 + 2 \text{NaNO}_3 \][/tex]
We see that 1 mole of [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex] produces 2 moles of [tex]\( \text{NaNO}_3 \)[/tex]. Thus, we calculate the moles of [tex]\( \text{NaNO}_3 \)[/tex] produced as follows:
[tex]\[ \text{Moles of } \text{NaNO}_3 = 2 \times \text{Moles of } \text{Na}_2\text{CrO}_4 = 2 \times 1.5617283950617284 = 3.123456790123457 \, \text{mol} \][/tex]
### Final Answer:
The number of moles of [tex]\( \text{NaNO}_3 \)[/tex] produced from the complete reaction of [tex]\( 253 \, \text{g} \, \text{Na}_2\text{CrO}_4 \)[/tex] is:
[tex]\[ 3.123456790123457 \, \text{mol} \][/tex]
### Step 1: Determine the Molar Mass of [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex]
Given:
- Sodium (Na): [tex]\( 23 \, \text{g/mol} \)[/tex]
- Chromium (Cr): [tex]\( 52 \, \text{g/mol} \)[/tex]
- Oxygen (O): [tex]\( 16 \, \text{g/mol} \)[/tex]
Thus, the molar mass of [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex] is calculated as follows:
[tex]\[ \text{Molar Mass of } \text{Na}_2\text{CrO}_4 = 2 \times 23 + 52 + 4 \times 16 = 46 + 52 + 64 = 162 \, \text{g/mol} \][/tex]
### Step 2: Calculate the Number of Moles of [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex]
Now, we use the molar mass to find the moles of [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex]:
[tex]\[ \text{Moles of } \text{Na}_2\text{CrO}_4 = \frac{\text{Mass of } \text{Na}_2\text{CrO}_4}{\text{Molar Mass of } \text{Na}_2\text{CrO}_4} = \frac{253 \, \text{g}}{162 \, \text{g/mol}} = 1.5617283950617284 \, \text{mol} \][/tex]
### Step 3: Use the Reaction Stoichiometry to Find the Moles of [tex]\( \text{NaNO}_3 \)[/tex]
From the balanced chemical equation:
[tex]\[ \text{Pb}\left(\text{NO}_3\right)_2 + \text{Na}_2\text{CrO}_4 \rightarrow \text{PbCrO}_4 + 2 \text{NaNO}_3 \][/tex]
We see that 1 mole of [tex]\( \text{Na}_2\text{CrO}_4 \)[/tex] produces 2 moles of [tex]\( \text{NaNO}_3 \)[/tex]. Thus, we calculate the moles of [tex]\( \text{NaNO}_3 \)[/tex] produced as follows:
[tex]\[ \text{Moles of } \text{NaNO}_3 = 2 \times \text{Moles of } \text{Na}_2\text{CrO}_4 = 2 \times 1.5617283950617284 = 3.123456790123457 \, \text{mol} \][/tex]
### Final Answer:
The number of moles of [tex]\( \text{NaNO}_3 \)[/tex] produced from the complete reaction of [tex]\( 253 \, \text{g} \, \text{Na}_2\text{CrO}_4 \)[/tex] is:
[tex]\[ 3.123456790123457 \, \text{mol} \][/tex]
