Question

name santiago escobar periodintro to stoichiometry - grams to grams practice$ce{2 nan3 -> 2 na + 3 n2}$1) given 0.38 grams of $ce{n2}$, how many grams of $ce{nan3}$ are needed?2) given 5.86 grams of $ce{nan3}$, how many grams of $ce{na}$ are produced?$ce{2 fe2o3 + 3 c -> 4 fe + 3 co2}$3) how many grams of $ce{fe}$ will be produced if 39.64 grams of $ce{co2}$ are used?

Explanation:

Step1: Calcular masa molar de $\text{N}_2$

Masa molar de $\text{N}_2 = 2 \times 14.01 = 28.02 \, \text{g/mol}$

Step2: Convertir gramos de $\text{N}_2$ a moles

$$n(\text{N}_2) = \frac{0.38 \, \text{g}}{28.02 \, \text{g/mol}} \approx 0.0136 \, \text{mol}$$

Step3: Relacionar moles de $\text{N}_2$ a $\text{NaN}_3$

De la reacción: $\frac{n(\text{NaN}_3)}{n(\text{N}_2)} = \frac{2}{3}$
$$n(\text{NaN}_3) = 0.0136 \, \text{mol} \times \frac{2}{3} \approx 0.00907 \, \text{mol}$$

Step4: Calcular masa molar de $\text{NaN}_3$

Masa molar de $\text{NaN}_3 = 22.99 + 3 \times 14.01 = 65.02 \, \text{g/mol}$

Step5: Convertir moles de $\text{NaN}_3$ a gramos

$$m(\text{NaN}_3) = 0.00907 \, \text{mol} \times 65.02 \, \text{g/mol} \approx 0.59 \, \text{g}$$

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Step6: Convertir gramos de $\text{NaN}_3$ a moles

$$n(\text{NaN}_3) = \frac{5.86 \, \text{g}}{65.02 \, \text{g/mol}} \approx 0.0901 \, \text{mol}$$

Step7: Relacionar moles de $\text{NaN}_3$ a $\text{Na}$

De la reacción: $\frac{n(\text{Na})}{n(\text{NaN}_3)} = \frac{2}{2} = 1$
$$n(\text{Na}) = 0.0901 \, \text{mol} \times 1 = 0.0901 \, \text{mol}$$

Step8: Calcular masa de $\text{Na}$ (masa molar = 22.99 g/mol)

$$m(\text{Na}) = 0.0901 \, \text{mol} \times 22.99 \, \text{g/mol} \approx 2.07 \, \text{g}$$

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Step9: Calcular masa molar de $\text{CO}_2$

Masa molar de $\text{CO}_2 = 12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol}$

Step10: Convertir gramos de $\text{CO}_2$ a moles

$$n(\text{CO}_2) = \frac{39.64 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.9007 \, \text{mol}$$

Step11: Relacionar moles de $\text{CO}_2$ a $\text{Fe}$

De la reacción: $\frac{n(\text{Fe})}{n(\text{CO}_2)} = \frac{4}{3}$
$$n(\text{Fe}) = 0.9007 \, \text{mol} \times \frac{4}{3} \approx 1.201 \, \text{mol}$$

Step12: Calcular masa de $\text{Fe}$ (masa molar = 55.85 g/mol)

$$m(\text{Fe}) = 1.201 \, \text{mol} \times 55.85 \, \text{g/mol} \approx 67.1 \, \text{g}$$

Answer:

1. 0.59 grams of $\text{NaN}_3$
2. 2.07 grams of $\text{Na}$
3. 67.1 grams of $\text{Fe}$