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Answer:
To determine the limiting reactant, we need to compare the amount of product that can be produced from each reactant and identify which reactant produces the least amount of product. We will use stoichiometry to calculate the amount of product formed from each reactant.
1. For the first reaction:
[tex]\[ 4Al_2O_3 + 9Fe \rightarrow 3Fe_3O_4 + 8Al \][/tex]
First, we need to find the molar masses of the substances involved:
[tex]- \( \text{Molar mass of } Al_2O_3 = 2 \times \text{Molar mass of } Al + 3 \times \text{Molar mass of } O = 2 \times 26.98 \text{ g/mol} + 3 \times 16.00 \text{ g/mol} = 101.96 \text{ g/mol} \)[/tex]
[tex]- \( \text{Molar mass of } Fe = 55.85 \text{ g/mol} \)[/tex]
Next, we need to find the moles of each reactant:
[tex]\[ \text{Moles of } Al_2O_3 = \frac{25.4 \text{ g}}{101.96 \text{ g/mol}} = 0.2493 \text{ moles} \][/tex]
[tex]\[ \text{Moles of } Fe = \frac{10.2 \text{ g}}{55.85 \text{ g/mol}} = 0.1826 \text{ moles} \][/tex]
Now, we need to determine which reactant limits the reaction by calculating the moles of product formed from each reactant. Using the stoichiometry of the balanced chemical equation:
[tex]\[ \text{Moles of } Al = 8 \times \text{Moles of } Al_2O_3 = 8 \times 0.2493 = 1.9944 \text{ moles} \][/tex]
[tex]\[ \text{Moles of } Fe_3O_4 = 3 \times \text{Moles of } Al_2O_3 = 3 \times 0.2493 = 0.7479 \text{ moles} \][/tex]
The limiting reactant is the one that produces the least amount of product, which in this case is [tex]Fe_3O_4.[/tex] So, the limiting reactant is [tex]\( \text{Al}_2\text{O}_3 \).[/tex]
2. After a reaction was run, there were 12.7 grams of reactant B left over. Reactant B is the
- superfluous reactant
- limiting reactant
- excess reactant
- extra reactant
In this case, since there is reactant left over after the reaction, it means that it was not fully consumed. Therefore, it is an excess reactant.
3. For the second reaction:
[tex]\[ Al_2(SO_3)_3 + 6NaOH \rightarrow 3Na_2SO_3 + 2Al(OH)_3 \][/tex]
First, we need to find the molar masses of the substances involved:
[tex]- \( \text{Molar mass of } Al_2(SO_3)_3 = 2 \times \text{Molar mass of } Al + 3 \times \text{Molar mass of } S + 9 \times \text{Molar mass of } O = 2 \times 26.98 \text{ g/mol} + 3 \times 32.07 \text{ g/mol} + 9 \times 16.00 \text{ g/mol} = 342.16 \text{ g/mol} \)[/tex]
[tex]- \( \text{Molar mass of } NaOH = 22.99 \text{ g/mol} + 15.999 \text{ g/mol} + 1.008 \text{ g/mol} = 39.997 \text{ g/mol} \)[/tex]
Next, we need to find the moles of each reactant:
[tex]\[ \text{Moles of } Al_2(SO_3)_3 = \frac{10.0 \text{ g}}{342.16 \text{ g/mol}} = 0.0292 \text{ moles} \][/tex]
[tex]\[ \text{Moles of } NaOH = \frac{10.0 \text{ g}}{39.997 \text{ g/mol}} = 0.2500 \text{ moles} \][/tex]
From the balanced equation, we see that the stoichiometric ratio between [tex]Al_2(SO_3)_3[/tex] NaOH is 1:6. Thus, for every 1 mole [tex]Al_2(SO_3)_3[/tex], 6 moles of NaOH are required. Since NaOH is used up in a smaller quantity [tex]Al_2(SO_3)_3[/tex], it is the limiting reactant.
4. For the third reaction:
[tex]\[ 3Zn + 2MoO_3 \rightarrow Mo_2O_3 + 3ZnO \][/tex]
First, we need to find the molar masses of the substances involved:
[tex]- \( \text{Molar mass of } MoO_3 = 95.95 \text{ g/mol} + 3 \times 15.999 \text{ g/mol} = 167.94 \text{ g/mol} \)[/tex]
[tex]- \( \text{Molar mass of } Zn = 65.38 \text{ g/mol} \)[/tex]
Next, we need to find the moles of each reactant:
[tex]\[ \text{Moles of } MoO_3 = \frac{20.0 \text{ g}}{167.94 \text{ g/mol}} = 0.1190 \text{ moles} \][/tex]
[tex]\[ \text{Moles of } Zn = \frac{10.0 \text{ g}}{65.38 \text{ g/mol}} = 0.1530 \text{ moles} \][/tex]
From the balanced equation, we see that the stoichiometric ratio between [tex]MoO_3[/tex] and Zn is 2:3. Thus, for every 2 moles of [tex]MoO_3[/tex], 3 moles of Zn are required. Since Zn is used up in a smaller quantity than [tex]MoO_3[/tex], it is the limiting reactant.
5. For the fourth reaction:
[tex]\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \][/tex]
First, we need to find the molar masses of the substances involved:
[tex]- \( \text{Molar mass of } C_3H_8 = 3 \times 12.01 \text{ g/mol} + 8 \times 1.008 \text{ g/mol} = 44.10 \text{ g/mol} \)[/tex]
[tex]- \( \text{Molar mass of } O_2 = 2 \times 15.999 \text{ g/mol} = 32.00 \text{ g/mol} \)[/tex]
Next, we need to find the moles of each reactant:
[tex]\[ \text{Moles of } C_3H_8 = \frac{14.8 \text{ g}}[/tex]
[tex]{44.10 \text{ g/mol}} = 0.3350 \text{ moles} \][/tex]
[tex]\[ \text{Moles of } O_2 = \frac{3.44 \text{ g}}{32.00 \text{ g/mol}} = 0.1075 \text{ moles} \][/tex]
From the balanced equation, we see that the stoichiometric ratio between[tex]C_3H_8 and O_2 is 1:5.[/tex] Thus, for every 1 mole of [tex]C_3H_8[/tex][tex], 5 moles of O_2[/tex]
are required. Since [tex]O_2[/tex] Is used up in a smaller quantity [tex]C_3H_8[/tex], it is the limiting reactant.
