Unit: Stoichiometry Answer key “Mole-Mole Calculations” - WS #1 ... - Free Printable

Here are the solutions to the stoichiometry problems shown in the worksheet. I have worked through each calculation step-by-step to ensure accuracy.

Problem 1

Part A: Potassium and Oxygen
* Reaction: $4 K + O_2 \rightarrow 2 K_2O$
* Given: 1.52 moles of Potassium ($K$).
* Goal: Find moles of Potassium Oxide ($K_2O$).
* Ratio: From the balanced equation, 4 moles of $K$ produce 2 moles of $K_2O$.
* Calculation:
$$1.52 \text{ mol } K \times \frac{2 \text{ mol } K_2O}{4 \text{ mol } K} = 0.76 \text{ mol } K_2O$$

Part B: Copper and Sulfur
* Reaction: $2 Cu + S \rightarrow Cu_2S$
* Given: 0.25 moles of Copper(I) sulfide ($Cu_2S$).
* Goal: Find moles of Copper ($Cu$) needed.
* Ratio: From the balanced equation, 1 mole of $Cu_2S$ requires 2 moles of $Cu$.
* Calculation:
$$0.25 \text{ mol } Cu_2S \times \frac{2 \text{ mol } Cu}{1 \text{ mol } Cu_2S} = 0.50 \text{ mol } Cu$$

Problem 2

Sodium and Sulfuric Acid
* Reaction: The equation needs balancing first.
Unbalanced: $\_\_ Na + \_\_ H_2(SO_4) \rightarrow \_\_ Na_2(SO_4) + \_\_ H_2$
Balanced: $2 Na + 1 H_2SO_4 \rightarrow 1 Na_2SO_4 + 1 H_2$
* Given: 3.75 moles of Sodium Sulfate ($Na_2SO_4$).
* Goal: Find moles of Sodium ($Na$) needed.
* Ratio: From the balanced equation, 1 mole of $Na_2SO_4$ requires 2 moles of $Na$.
* Calculation:
$$3.75 \text{ mol } Na_2SO_4 \times \frac{2 \text{ mol } Na}{1 \text{ mol } Na_2SO_4} = 7.50 \text{ mol } Na$$

Problem 3

Iron(III) Oxide and Calcium Phosphate
* Reaction: The equation needs balancing.
Unbalanced: $\_\_ Fe_2O_3 + \_\_ Ca_3(PO_4)_2 \rightarrow \_\_ Fe(PO_4) + \_\_ CaO$
*Note: The product is likely Iron(III) phosphate, $FePO_4$, based on the charges.*
Balanced: $1 Fe_2O_3 + 1 Ca_3(PO_4)_2 \rightarrow 2 FePO_4 + 3 CaO$
* Given: 1.6 moles of Iron(III) oxide ($Fe_2O_3$).
* Goal: Find moles of Calcium Oxide ($CaO$) produced.
* Ratio: From the balanced equation, 1 mole of $Fe_2O_3$ produces 3 moles of $CaO$.
* Calculation:
$$1.6 \text{ mol } Fe_2O_3 \times \frac{3 \text{ mol } CaO}{1 \text{ mol } Fe_2O_3} = 4.8 \text{ mol } CaO$$

Problem 4

Magnesium Chlorate Decomposition
* Reaction: Write and balance the formula equation.
Word: magnesium chlorate $\rightarrow$ magnesium chloride + oxygen
Formula: $Mg(ClO_3)_2 \rightarrow MgCl_2 + O_2$
Balanced: $1 Mg(ClO_3)_2 \rightarrow 1 MgCl_2 + 3 O_2$
* Given: 5.12 moles of Magnesium Chlorate ($Mg(ClO_3)_2$).
* Goal: Find moles of Oxygen ($O_2$) produced.
* Ratio: From the balanced equation, 1 mole of reactant produces 3 moles of $O_2$.
* Calculation:
$$5.12 \text{ mol } Mg(ClO_3)_2 \times \frac{3 \text{ mol } O_2}{1 \text{ mol } Mg(ClO_3)_2} = 15.36 \text{ mol } O_2$$
*(Rounding to 3 significant figures gives 15.4 mol)*

Problem 5

Chromium and Lead(IV) Nitrate
* Reaction: Write and balance the formula equation.
Word: chromium + lead (IV) nitrate $\rightarrow$ chromium (III) nitrate + lead
Formula: $Cr + Pb(NO_3)_4 \rightarrow Cr(NO_3)_3 + Pb$
To balance the nitrates ($NO_3$), we find the common multiple of 4 and 3, which is 12.
Balanced: $4 Cr + 3 Pb(NO_3)_4 \rightarrow 4 Cr(NO_3)_3 + 3 Pb$
* Given: 0.85 moles of Lead(IV) nitrate ($Pb(NO_3)_4$).
* Goal: Find moles of Chromium(III) nitrate ($Cr(NO_3)_3$) produced.
* Ratio: From the balanced equation, 3 moles of $Pb(NO_3)_4$ produce 4 moles of $Cr(NO_3)_3$.
* Calculation:
$$0.85 \text{ mol } Pb(NO_3)_4 \times \frac{4 \text{ mol } Cr(NO_3)_3}{3 \text{ mol } Pb(NO_3)_4} \approx 1.133 \text{ mol } Cr(NO_3)_3$$
*(Rounding to 2 significant figures based on "0.85" gives 1.1 mol)*

──────────────────────────────────────

Final Answer:
1. Potassium/Oxygen: 0.76 mol $K_2O$
Copper/Sulfur: 0.50 mol $Cu$
2. 7.50 mol $Na$
3. 4.8 mol $CaO$
4. 15.4 mol $O_2$
5. 1.1 mol $Cr(NO_3)_3$