the chemical equation below shows the decomposition of nitrogen triiodide (ni3) into nitrogen (n2) and…

the chemical equation below shows the decomposition of nitrogen triiodide (ni3) into nitrogen (n2) and iodine (i2). 2ni3 → n2 + 3i2 the molar mass of ni3 is 394.71 g/mol. how many moles of i2 will form 3.58 g of ni3? 0.00907 moles 0.00940 moles 0.0136 moles 0.0212 moles

the chemical equation below shows the decomposition of nitrogen triiodide (ni3) into nitrogen (n2) and iodine (i2). 2ni3 → n2 + 3i2 the molar mass of ni3 is 394.71 g/mol. how many moles of i2 will form 3.58 g of ni3? 0.00907 moles 0.00940 moles 0.0136 moles 0.0212 moles

Answer

Explanation:

Step1: Calculate moles of NI3

Use the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass and $M$ is the molar - mass. Given $m = 3.58\ g$ and $M=394.71\ g/mol$, so $n_{NI3}=\frac{3.58\ g}{394.71\ g/mol}\approx0.00907\ mol$.

Step2: Use mole - ratio from the balanced equation

The balanced equation is $2NI_3\rightarrow N_2 + 3I_2$. The mole - ratio of $NI_3$ to $I_2$ is $2:3$. Let the number of moles of $I_2$ be $n_{I2}$. Then $\frac{n_{I2}}{n_{NI3}}=\frac{3}{2}$, so $n_{I2}=\frac{3}{2}\times n_{NI3}$. Substitute $n_{NI3}=0.00907\ mol$ into the equation: $n_{I2}=\frac{3}{2}\times0.00907\ mol = 0.0136\ mol$.

Answer:

0.0136 moles