Question

how many grams of sodium iodide are required to react completely with 3.01×10²³ molecules of chlorine?
2 nai + cl₂ → i₂ + 2 nacl
37.5 g

1. g

74.9 g
1.50 g

Explanation:

Step1: Calculate moles of chlorine

Use Avogadro's number ($N_A = 6.02\times10^{23}\text{ molecules/mol}$). Moles of $Cl_2$, $n_{Cl_2}=\frac{3.01\times 10^{23}\text{ molecules}}{6.02\times 10^{23}\text{ molecules/mol}} = 0.5\text{ mol}$

Step2: Determine moles of sodium iodide

From the balanced chemical equation $2NaI + Cl_2
ightarrow I_2+2NaCl$, the mole - ratio of $NaI$ to $Cl_2$ is 2:1. So moles of $NaI$, $n_{NaI}=2\times n_{Cl_2}=2\times0.5\text{ mol}=1\text{ mol}$

Step3: Calculate mass of sodium iodide

The molar mass of $NaI$ is $M_{NaI}=22.99\text{ g/mol}+126.90\text{ g/mol}=149.89\text{ g/mol}$. Mass of $NaI$, $m_{NaI}=n_{NaI}\times M_{NaI}=1\text{ mol}\times149.89\text{ g/mol}\approx150\text{ g}$

Answer:

1. g