Question

stoichiometry - grams & moles calculations
**success criteria: i can calculate using molar mass from grams to moles or moles to grams.
directions: given the following, find the number of moles:

1. 30 grams of $ce{h3po4}$
2. 25 grams of hydrogen gas
3. 110 grams of $ce{nahco3}$
4. 1.1 grams of $ce{fecl3}$
5. 987 grams of $ce{ra(oh)2}$
6. 564 grams of copper
7. 12.3 grams of $ce{co2}$
8. 89 grams of $ce{pb(ch3coo)4}$

given the following, find the number of grams:

1. 4 moles $ce{cu(cn)2}$
2. 5.6 moles of $ce{c6h6}$
3. 21.3 moles of $ce{baco3}$
4. 1.2 moles of $ce{(nh4)3po3}$
5. $9.3 \times 10^{3-}$ moles of $ce{smo}$
6. 6.6 moles of $ce{zno}$
7. 5.4 moles of $ce{k2so4}$

Explanation:

Step1: Calculate molar mass of $\text{H}_3\text{PO}_4$

Molar mass = $3(1.008) + 30.97 + 4(16.00) = 97.994\ \text{g/mol}$

Step2: Moles of $\text{H}_3\text{PO}_4$

$\text{Moles} = \frac{30}{97.994} \approx 0.306$

Step3: Molar mass of $\text{H}_2$

Molar mass = $2(1.008) = 2.016\ \text{g/mol}$

Step4: Moles of $\text{H}_2$

$\text{Moles} = \frac{25}{2.016} \approx 12.40$

Step5: Molar mass of $\text{NaHCO}_3$

Molar mass = $22.99 + 1.008 + 12.01 + 3(16.00) = 84.008\ \text{g/mol}$

Step6: Moles of $\text{NaHCO}_3$

$\text{Moles} = \frac{110}{84.008} \approx 1.31$

Step7: Molar mass of $\text{FeCl}_3$

Molar mass = $55.85 + 3(35.45) = 162.2\ \text{g/mol}$

Step8: Moles of $\text{FeCl}_3$

$\text{Moles} = \frac{1.1}{162.2} \approx 0.00678$

Step9: Molar mass of $\text{Ra(OH)}_2$

Molar mass = $226.03 + 2(16.00 + 1.008) = 259.046\ \text{g/mol}$

Step10: Moles of $\text{Ra(OH)}_2$

$\text{Moles} = \frac{987}{259.046} \approx 3.81$

Step11: Molar mass of $\text{Cu}$

Molar mass = $63.55\ \text{g/mol}$

Step12: Moles of $\text{Cu}$

$\text{Moles} = \frac{564}{63.55} \approx 8.88$

Step13: Molar mass of $\text{CO}_2$

Molar mass = $12.01 + 2(16.00) = 44.01\ \text{g/mol}$

Step14: Moles of $\text{CO}_2$

$\text{Moles} = \frac{12.3}{44.01} \approx 0.279$

Step15: Molar mass of $\text{Pb(CH}_3\text{COO)}_4$

Molar mass = $207.2 + 4(2(12.01)+3(1.008)+2(16.00)) = 443.344\ \text{g/mol}$

Step16: Moles of $\text{Pb(CH}_3\text{COO)}_4$

$\text{Moles} = \frac{89}{443.344} \approx 0.201$

Step17: Molar mass of $\text{Cu(CN)}_2$

Molar mass = $63.55 + 2(12.01 + 14.01) = 115.59\ \text{g/mol}$

Step18: Grams of $\text{Cu(CN)}_2$

$\text{Grams} = 4 \times 115.59 = 462.36$

Step19: Molar mass of $\text{C}_6\text{H}_6$

Molar mass = $6(12.01) + 6(1.008) = 78.108\ \text{g/mol}$

Step20: Grams of $\text{C}_6\text{H}_6$

$\text{Grams} = 5.6 \times 78.108 = 437.40$

Step21: Molar mass of $\text{BaCO}_3$

Molar mass = $137.33 + 12.01 + 3(16.00) = 197.34\ \text{g/mol}$

Step22: Grams of $\text{BaCO}_3$

$\text{Grams} = 21.3 \times 197.34 = 4203.34$

Step23: Molar mass of $(\text{NH}_4)_3\text{PO}_3$

Molar mass = $3(14.01+4(1.008)) + 30.97 + 3(16.00) = 133.09\ \text{g/mol}$

Step24: Grams of $(\text{NH}_4)_3\text{PO}_3$

$\text{Grams} = 1.2 \times 133.09 = 159.71$

Step25: Molar mass of $\text{SmO}$

Molar mass = $150.36 + 16.00 = 166.36\ \text{g/mol}$

Step26: Grams of $\text{SmO}$

$\text{Grams} = 9.3 \times 10^{-3} \times 166.36 \approx 1.55$

Step27: Molar mass of $\text{ZnO}$

Molar mass = $65.38 + 16.00 = 81.38\ \text{g/mol}$

Step28: Grams of $\text{ZnO}$

$\text{Grams} = 6.6 \times 81.38 = 537.11$

Step29: Molar mass of $\text{K}_2\text{SO}_4$

Molar mass = $2(39.10) + 32.07 + 4(16.00) = 174.27\ \text{g/mol}$

Step30: Grams of $\text{K}_2\text{SO}_4$

$\text{Grams} = 5.4 \times 174.27 = 941.06$

Answer:

1. 0.306 moles
2. 12.40 moles
3. 1.31 moles
4. 0.00678 moles
5. 3.81 moles
6. 8.88 moles
7. 0.279 moles
8. 0.201 moles
9. 462.36 grams
10. 437.40 grams
11. 4203.34 grams
12. 159.71 grams
13. 1.55 grams
14. 537.11 grams
15. 941.06 grams