Question

The chemical equation below shows the decomposition of nitrogen triiodide (NI3) into nitrogen (N2) and iodine (I2). 2NI3 N2 + 3I2 The molar mass of I2 is 253.80 g/mol, and the molar mass of NI3 is 394.71 g/mol. How many moles of I2 will form 3.58 g of NI3?

Answer 1

On the basis of 3.58 g of NI3, it will obtained 0.013605 mole of I2

Step-by-step explanation:

We have 3.58 g or NI3, taking into account the molar relation between NI3 and I2, established in the reaction, we can calculate the mole of I2 in the next way:

3.58g NI3 * (1 mole / 394.71 g NI3) * (3 mole I2 / 2 mole NI3) = 0.013605 mole I2.

Answer 2

`\boxed{\text{0.0136 mol I}_(2)}`

Step-by-step explanation:

We know we will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

M_r: 394.71 253.80

2NI₃ ⟶ N₂ + 3I₂

Mass/g: 3.58

1. Calculate the moles of NI₃

`\text{Moles of NI$_(3)$} = \text{3.58 g NI}_(3) * \frac{\text{1 mol NI}_(3)}{\text{394.71 g NII}_(3)} = 9.070 * 10^(-3) \text{ mol NI}_(3)`

2. Use the molar ratio of I₂:NI₃ to calculate the moles of I₂

`\text{Moles of I}_(2) = 9.070 * 10^(-3)\text{ mol NI}_(3)* \frac{\text{3 mol I}_(2)}{\text{2 mol NI$_(3)$}}=\textbf{0.0136 mol I}_{\mathbf{2}}\\\\\text{The reaction will form }\boxed{\textbf{0.0136 mol I}_{\mathbf{2}}}`