Free Printable Mass to Mole Stoichiometry Worksheets - Free Printable
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Step-by-step solution for: Free Printable Mass to Mole Stoichiometry Worksheets
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Step-by-step solution for: Free Printable Mass to Mole Stoichiometry Worksheets
Here are the step-by-step solutions for each problem on the worksheet.
a. If $15 \text{ g}$ of $\text{N}_2\text{O}_4$ was produced, how many moles of $\text{O}_2$ were required?
1. Find the Molar Mass of $\text{N}_2\text{O}_4$:
* Nitrogen (N) = $14.01 \text{ g/mol}$
* Oxygen (O) = $16.00 \text{ g/mol}$
* $\text{N}_2\text{O}_4 = (2 \times 14.01) + (4 \times 16.00) = 28.02 + 64.00 = 92.02 \text{ g/mol}$.
2. Convert Grams to Moles:
* $\text{Moles of N}_2\text{O}_4 = \frac{15 \text{ g}}{92.02 \text{ g/mol}} \approx 0.163 \text{ mol}$.
3. Use the Mole Ratio:
* According to the equation, $1 \text{ mole of N}_2\text{O}_4$ requires $2 \text{ moles of O}_2$.
* $\text{Moles of O}_2 = 0.163 \text{ mol} \times 2 = 0.326 \text{ mol}$.
b. If $4 \times 10^{-3} \text{ moles}$ of oxygen reacted, how many grams of $\text{N}_2$ were needed?
1. Use the Mole Ratio:
* The ratio is $1 \text{ mole N}_2$ for every $2 \text{ moles O}_2$.
* $\text{Moles of N}_2 = \frac{4 \times 10^{-3} \text{ mol O}_2}{2} = 2 \times 10^{-3} \text{ mol N}_2$ (or $0.002 \text{ mol}$).
2. Convert Moles to Grams:
* Molar Mass of $\text{N}_2 = 2 \times 14.01 = 28.02 \text{ g/mol}$.
* $\text{Mass} = 0.002 \text{ mol} \times 28.02 \text{ g/mol} = 0.05604 \text{ g}$.
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a. What is the mass of potassium nitrate produced when $2.04 \text{ moles}$ of potassium phosphate react?
1. Use the Mole Ratio:
* The equation shows $1 \text{ mole of K}_3\text{PO}_4$ produces $3 \text{ moles of KNO}_3$.
* $\text{Moles of KNO}_3 = 2.04 \text{ mol} \times 3 = 6.12 \text{ mol}$.
2. Find the Molar Mass of $\text{KNO}_3$:
* Potassium (K) = $39.10$, Nitrogen (N) = $14.01$, Oxygen (O) = $16.00$.
* $\text{KNO}_3 = 39.10 + 14.01 + (3 \times 16.00) = 101.11 \text{ g/mol}$.
3. Calculate Mass:
* $\text{Mass} = 6.12 \text{ mol} \times 101.11 \text{ g/mol} = 618.79 \text{ g}$.
b. If $5.8 \text{ g}$ of aluminum phosphate are formed, how many moles of aluminum nitrate reacted?
1. Find the Molar Mass of $\text{AlPO}_4$:
* Aluminum (Al) = $26.98$, Phosphorus (P) = $30.97$, Oxygen (O) = $16.00$.
* $\text{AlPO}_4 = 26.98 + 30.97 + (4 \times 16.00) = 121.95 \text{ g/mol}$.
2. Convert Grams to Moles:
* $\text{Moles of AlPO}_4 = \frac{5.8 \text{ g}}{121.95 \text{ g/mol}} \approx 0.0476 \text{ mol}$.
3. Use the Mole Ratio:
* The ratio between $\text{Al(NO}_3)_3$ and $\text{AlPO}_4$ is $1:1$.
* Therefore, moles of $\text{Al(NO}_3)_3$ reacted = $0.0476 \text{ mol}$.
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a. If you have $5.5 \text{ mol}$ of $\text{CaC}_2$, how much $\text{C}_2\text{H}_2$ do you get?
1. Use the Mole Ratio:
* The coefficients for $\text{CaC}_2$ and $\text{C}_2\text{H}_2$ are both $1$. The ratio is $1:1$.
* Therefore, $5.5 \text{ mol of CaC}_2$ produces $5.5 \text{ mol of C}_2\text{H}_2$.
b. How many moles of water are needed when $65 \text{ g}$ of $\text{CaC}_2$ have reacted?
1. Find the Molar Mass of $\text{CaC}_2$:
* Calcium (Ca) = $40.08$, Carbon (C) = $12.01$.
* $\text{CaC}_2 = 40.08 + (2 \times 12.01) = 64.10 \text{ g/mol}$.
2. Convert Grams to Moles:
* $\text{Moles of CaC}_2 = \frac{65 \text{ g}}{64.10 \text{ g/mol}} \approx 1.014 \text{ mol}$.
3. Use the Mole Ratio:
* The equation requires $2 \text{ moles of H}_2\text{O}$ for every $1 \text{ mole of CaC}_2$.
* $\text{Moles of H}_2\text{O} = 1.014 \text{ mol} \times 2 = 2.028 \text{ mol}$.
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What mass of oxygen gas is consumed in a reaction that produces $4.6 \text{ mol SO}_2$?
1. Use the Mole Ratio:
* Looking at the coefficients: $3 \text{ moles of O}_2$ produce $2 \text{ moles of SO}_2$.
* $\text{Moles of O}_2 = 4.6 \text{ mol SO}_2 \times \frac{3 \text{ mol O}_2}{2 \text{ mol SO}_2}$.
* $\text{Moles of O}_2 = 4.6 \times 1.5 = 6.9 \text{ mol}$.
2. Find the Molar Mass of $\text{O}_2$:
* $\text{O}_2 = 2 \times 16.00 = 32.00 \text{ g/mol}$.
3. Calculate Mass:
* $\text{Mass} = 6.9 \text{ mol} \times 32.00 \text{ g/mol} = 220.8 \text{ g}$.
Final Answer:
1a. 0.33 mol O₂
1b. 0.056 g N₂
2a. 619 g KNO₃
2b. 0.048 mol Al(NO₃)₃
3a. 5.5 mol C₂H₂
3b. 2.0 mol H₂O
4. 221 g O₂
1. Reaction: $\text{N}_2 + 2 \text{O}_2 \rightarrow \text{N}_2\text{O}_4$
a. If $15 \text{ g}$ of $\text{N}_2\text{O}_4$ was produced, how many moles of $\text{O}_2$ were required?
1. Find the Molar Mass of $\text{N}_2\text{O}_4$:
* Nitrogen (N) = $14.01 \text{ g/mol}$
* Oxygen (O) = $16.00 \text{ g/mol}$
* $\text{N}_2\text{O}_4 = (2 \times 14.01) + (4 \times 16.00) = 28.02 + 64.00 = 92.02 \text{ g/mol}$.
2. Convert Grams to Moles:
* $\text{Moles of N}_2\text{O}_4 = \frac{15 \text{ g}}{92.02 \text{ g/mol}} \approx 0.163 \text{ mol}$.
3. Use the Mole Ratio:
* According to the equation, $1 \text{ mole of N}_2\text{O}_4$ requires $2 \text{ moles of O}_2$.
* $\text{Moles of O}_2 = 0.163 \text{ mol} \times 2 = 0.326 \text{ mol}$.
b. If $4 \times 10^{-3} \text{ moles}$ of oxygen reacted, how many grams of $\text{N}_2$ were needed?
1. Use the Mole Ratio:
* The ratio is $1 \text{ mole N}_2$ for every $2 \text{ moles O}_2$.
* $\text{Moles of N}_2 = \frac{4 \times 10^{-3} \text{ mol O}_2}{2} = 2 \times 10^{-3} \text{ mol N}_2$ (or $0.002 \text{ mol}$).
2. Convert Moles to Grams:
* Molar Mass of $\text{N}_2 = 2 \times 14.01 = 28.02 \text{ g/mol}$.
* $\text{Mass} = 0.002 \text{ mol} \times 28.02 \text{ g/mol} = 0.05604 \text{ g}$.
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2. Reaction: $\text{K}_3\text{PO}_4 + \text{Al(NO}_3)_3 \rightarrow 3 \text{KNO}_3 + \text{AlPO}_4$
a. What is the mass of potassium nitrate produced when $2.04 \text{ moles}$ of potassium phosphate react?
1. Use the Mole Ratio:
* The equation shows $1 \text{ mole of K}_3\text{PO}_4$ produces $3 \text{ moles of KNO}_3$.
* $\text{Moles of KNO}_3 = 2.04 \text{ mol} \times 3 = 6.12 \text{ mol}$.
2. Find the Molar Mass of $\text{KNO}_3$:
* Potassium (K) = $39.10$, Nitrogen (N) = $14.01$, Oxygen (O) = $16.00$.
* $\text{KNO}_3 = 39.10 + 14.01 + (3 \times 16.00) = 101.11 \text{ g/mol}$.
3. Calculate Mass:
* $\text{Mass} = 6.12 \text{ mol} \times 101.11 \text{ g/mol} = 618.79 \text{ g}$.
b. If $5.8 \text{ g}$ of aluminum phosphate are formed, how many moles of aluminum nitrate reacted?
1. Find the Molar Mass of $\text{AlPO}_4$:
* Aluminum (Al) = $26.98$, Phosphorus (P) = $30.97$, Oxygen (O) = $16.00$.
* $\text{AlPO}_4 = 26.98 + 30.97 + (4 \times 16.00) = 121.95 \text{ g/mol}$.
2. Convert Grams to Moles:
* $\text{Moles of AlPO}_4 = \frac{5.8 \text{ g}}{121.95 \text{ g/mol}} \approx 0.0476 \text{ mol}$.
3. Use the Mole Ratio:
* The ratio between $\text{Al(NO}_3)_3$ and $\text{AlPO}_4$ is $1:1$.
* Therefore, moles of $\text{Al(NO}_3)_3$ reacted = $0.0476 \text{ mol}$.
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3. Reaction: $\text{CaC}_2 + 2 \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{C}_2\text{H}_2$
a. If you have $5.5 \text{ mol}$ of $\text{CaC}_2$, how much $\text{C}_2\text{H}_2$ do you get?
1. Use the Mole Ratio:
* The coefficients for $\text{CaC}_2$ and $\text{C}_2\text{H}_2$ are both $1$. The ratio is $1:1$.
* Therefore, $5.5 \text{ mol of CaC}_2$ produces $5.5 \text{ mol of C}_2\text{H}_2$.
b. How many moles of water are needed when $65 \text{ g}$ of $\text{CaC}_2$ have reacted?
1. Find the Molar Mass of $\text{CaC}_2$:
* Calcium (Ca) = $40.08$, Carbon (C) = $12.01$.
* $\text{CaC}_2 = 40.08 + (2 \times 12.01) = 64.10 \text{ g/mol}$.
2. Convert Grams to Moles:
* $\text{Moles of CaC}_2 = \frac{65 \text{ g}}{64.10 \text{ g/mol}} \approx 1.014 \text{ mol}$.
3. Use the Mole Ratio:
* The equation requires $2 \text{ moles of H}_2\text{O}$ for every $1 \text{ mole of CaC}_2$.
* $\text{Moles of H}_2\text{O} = 1.014 \text{ mol} \times 2 = 2.028 \text{ mol}$.
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4. Reaction: $2 \text{H}_2\text{S (g)} + 3 \text{O}_2 \text{(g)} \rightarrow 2 \text{SO}_2 \text{(g)} + 2 \text{H}_2\text{O (g)}$
What mass of oxygen gas is consumed in a reaction that produces $4.6 \text{ mol SO}_2$?
1. Use the Mole Ratio:
* Looking at the coefficients: $3 \text{ moles of O}_2$ produce $2 \text{ moles of SO}_2$.
* $\text{Moles of O}_2 = 4.6 \text{ mol SO}_2 \times \frac{3 \text{ mol O}_2}{2 \text{ mol SO}_2}$.
* $\text{Moles of O}_2 = 4.6 \times 1.5 = 6.9 \text{ mol}$.
2. Find the Molar Mass of $\text{O}_2$:
* $\text{O}_2 = 2 \times 16.00 = 32.00 \text{ g/mol}$.
3. Calculate Mass:
* $\text{Mass} = 6.9 \text{ mol} \times 32.00 \text{ g/mol} = 220.8 \text{ g}$.
Final Answer:
1a. 0.33 mol O₂
1b. 0.056 g N₂
2a. 619 g KNO₃
2b. 0.048 mol Al(NO₃)₃
3a. 5.5 mol C₂H₂
3b. 2.0 mol H₂O
4. 221 g O₂
Parent Tip: Review the logic above to help your child master the concept of mass to moles worksheet.
