Question

1. $\boldsymbol{6 na_{(s)} + fe_{2}o_{3 (s)}

ightarrow 3 na_{2}o_{(s)} + 2 fe_{(s)}}$if 100.0 grams of na and 100.0 grams of $fe_{2}o_{3}$ are used in this reaction determine the:a. limiting reactant _______________b. the excess reactant _______________5. $\boldsymbol{agno_{3 (aq)} + kbr_{(aq)}
ightarrow agbr_{(s)} + kno_{3 (aq)}}$a reaction combines 64.81 grams of $agno_{3}$ with 92.67 grams of kbr.a. how much agbr is formed? _____________ich reactant is limiting? _____________ which reactant is in e

Explanation:

Step1: Calculate moles of Na

Molar mass of Na = $22.99\ \text{g/mol}$
$\text{Moles of Na} = \frac{100.0\ \text{g}}{22.99\ \text{g/mol}} \approx 4.35\ \text{mol}$

Step2: Calculate moles of $\text{Fe}_2\text{O}_3$

Molar mass of $\text{Fe}_2\text{O}_3 = (2\times55.85)+(3\times16.00)=159.70\ \text{g/mol}$
$\text{Moles of }\text{Fe}_2\text{O}_3 = \frac{100.0\ \text{g}}{159.70\ \text{g/mol}} \approx 0.626\ \text{mol}$

Step3: Find Fe from Na (reaction ratio)

Reaction ratio: $\frac{2\ \text{mol Fe}}{6\ \text{mol Na}}$
$\text{Moles of Fe from Na} = 4.35\ \text{mol Na} \times \frac{2\ \text{mol Fe}}{6\ \text{mol Na}} \approx 1.45\ \text{mol}$

Step4: Find Fe from $\text{Fe}_2\text{O}_3$ (reaction ratio)

Reaction ratio: $\frac{2\ \text{mol Fe}}{1\ \text{mol }\text{Fe}_2\text{O}_3}$
$\text{Moles of Fe from }\text{Fe}_2\text{O}_3 = 0.626\ \text{mol }\text{Fe}_2\text{O}_3 \times \frac{2\ \text{mol Fe}}{1\ \text{mol }\text{Fe}_2\text{O}_3} \approx 1.25\ \text{mol}$

Step5: Identify limiting reactant (Q4a)

The reactant producing less product is limiting. $\text{Fe}_2\text{O}_3$ produces less Fe, so it is limiting.

Step6: Identify excess reactant (Q4b)

The remaining reactant is in excess: Na.

Step7: Calculate moles of $\text{AgNO}_3$ (Q5)

Molar mass of $\text{AgNO}_3 = 107.87+14.01+(3\times16.00)=169.88\ \text{g/mol}$
$\text{Moles of }\text{AgNO}_3 = \frac{64.81\ \text{g}}{169.88\ \text{g/mol}} \approx 0.3815\ \text{mol}$

Step8: Calculate moles of KBr (Q5)

Molar mass of KBr = $39.10+79.90=119.00\ \text{g/mol}$
$\text{Moles of KBr} = \frac{92.67\ \text{g}}{119.00\ \text{g/mol}} \approx 0.7788\ \text{mol}$

Step9: Find limiting reactant (Q5)

Reaction ratio is 1:1. $\text{AgNO}_3$ has fewer moles, so it is limiting.

Step10: Calculate mass of AgBr (Q5a)

Molar mass of AgBr = $107.87+79.90=187.77\ \text{g/mol}$
Moles of AgBr = moles of limiting $\text{AgNO}_3 = 0.3815\ \text{mol}$
$\text{Mass of AgBr} = 0.3815\ \text{mol} \times 187.77\ \text{g/mol} \approx 71.6\ \text{g}$

Step11: Identify excess reactant (Q5)

The remaining reactant is in excess: KBr.

Answer:

1. a. $\text{Fe}_2\text{O}_3$

b. $\text{Na}$

1. a. $71.6\ \text{grams}$

Limiting reactant: $\text{AgNO}_3$
Excess reactant: $\text{KBr}$