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

Answer:

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

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

Answer:

$\boxed{\text{0.0136 mol I}_{2}}$

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

$\text{Moles of NI _{3} } = \text{3.58 g NI}_{3} \times \dfrac{\text{1 mol NI}_{3}}{\text{394.71 g NII}_{3}} = 9.070 \times 10^{-3} \text{ mol NI}_{3}$

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

$\text{Moles of I}_{2} = 9.070 \times 10^{-3}\text{ mol NI}_{3}\times \dfrac{\text{3 mol I}_{2}}{\text{2 mol NI _{3} }}=\textbf{0.0136 mol I}_{\mathbf{2}}\\\\\text{The reaction will form }\boxed{\textbf{0.0136 mol I}_{\mathbf{2}}}$