Question

name santiago escobar
period 3b
intro to stoichiometry - grams to grams practice
$2\text{nan}_3 \
ightarrow 2\text{na} + 3\text{n}_2$

1. given 0.38 grams of $\text{n}_2$, how many grams of $\text{nan}_3$ are needed?

Explanation:

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

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

Step2: Calcular moles de $\text{N}_2$

$\text{Moles de } \text{N}_2 = \frac{0.38 \, \text{g}}{28.02 \, \frac{\text{g}}{\text{mol}}} \approx 0.01356 \, \text{mol}$

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

De la reacción: $\frac{\text{Moles de } \text{NaN}_3}{\text{Moles de } \text{N}_2} = \frac{2}{3}$
$\text{Moles de } \text{NaN}_3 = 0.01356 \, \text{mol} \times \frac{2}{3} \approx 0.00904 \, \text{mol}$

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

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

Step5: Calcular masa de $\text{NaN}_3$

$\text{Masa de } \text{NaN}_3 = 0.00904 \, \text{mol} \times 65.02 \, \frac{\text{g}}{\text{mol}} \approx 0.59 \, \text{g}$

Answer:

0.59 gramos de $\text{NaN}_3$