Pergunta
Problem 2: Consider the following reaction: 3NH_(4)NO_(3)+Na_(3)PO_(4)arrow (NH4)_(3)PO_(4)+3NaNO_(3) Given: 50.0 grams of ammonium nitrate and 60.0 grams of sodium phosphate. Find: Which of the reagents is the limiting reagent? NH_(4)NO_(3) Na_(3)PO_(4) (NH4)_(3)PO_(4) NaNO_(3)
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MayraMestre · Tutor por 5 anos
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To determine the limiting reagent, we need to calculate the number of moles of each reactant and compare them.
First, let's calculate the molar mass of each compound:
- NH_{4}NO_{3}
- Na_{3}PO_{4}
Now, let's calculate the number of moles of each reactant:
- Moles of NH_{4}NO_{3}
- Moles of Na_{3}PO_{4}
Next, we need to compare the mole ratio of the reactants to the stoichiometric ratio in the balanced chemical equation. The balanced chemical equation is:
3NH_{4}NO_{3} + Na_{3}PO_{4} \rightarrow (NH_{4})_{3}PO_{4} + 3NaNO_{3}
From the balanced equation, we can see that the stoichiometric ratio of NH_{4}NO_{3}
Now, let's compare the mole ratio of the reactants to the stoichiometric ratio:
- Mole ratio of NH_{4}NO_{3}
Since the mole ratio of NH_{4}NO_{3}
Therefore, the limiting reagent is Na_{3}PO_{4}
Answer: Na_{3}PO_{4}
First, let's calculate the molar mass of each compound:
- NH_{4}NO_{3}
: 14.01 + (4 \times 1.01) + 14.01 + (3 \times 16.00) = 80.05 \, \text{g/mol}
- Na_{3}PO_{4}
: (3 \times 22.99) + 30.97 + (4 \times 16.00) = 163.94 \, \text{g/mol}
Now, let's calculate the number of moles of each reactant:
- Moles of NH_{4}NO_{3}
: \frac{50.0 \, \text{g}}{80.05 \, \text{g/mol}} = 0.625 \, \text{mol}
- Moles of Na_{3}PO_{4}
: \frac{60.0 \, \text{g}}{163.94 \, \text{g/mol}} = 0.366 \, \text{mol}
Next, we need to compare the mole ratio of the reactants to the stoichiometric ratio in the balanced chemical equation. The balanced chemical equation is:
3NH_{4}NO_{3} + Na_{3}PO_{4} \rightarrow (NH_{4})_{3}PO_{4} + 3NaNO_{3}
From the balanced equation, we can see that the stoichiometric ratio of NH_{4}NO_{3}
to Na_{3}PO_{4}
is 3:1.
Now, let's compare the mole ratio of the reactants to the stoichiometric ratio:
- Mole ratio of NH_{4}NO_{3}
to Na_{3}PO_{4}
: \frac{0.625 \, \text{mol}}{0.366 \, \text{mol}} = 1.71
Since the mole ratio of NH_{4}NO_{3}
to Na_{3}PO_{4}
is greater than the stoichiometric ratio of 3:1, this means that there is an excess of NH_{4}NO_{3}
.
Therefore, the limiting reagent is Na_{3}PO_{4}
.
Answer: Na_{3}PO_{4}
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