Question

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

The Answer Is C. 0.0136

Answer 2

Answer : The number of moles of $I_2$ form will be, 0.01360 moles

Solution : Given,

Mass of $NI_3$ = 3.58 g

Molar mass of $NI_3$ = 394.71 g/mole

Molar mass of $I_2$ = 253.80 g/mole

First we to calculate the moles of moles of $NI_3$.

$\text{Moles of }NI_3=\frac{\text{Mass of }NI_3}{\text{Molar mass of }NI_3}=\frac{3.58g}{394.71g/mole}=9.069\times 10^{-3}moles$

Now we have to calculate the moles of $I_2$.

The balanced reaction is,

$2NI_3\rightarrow N_2+3I_2$

From the balanced reaction, we conclude that

As, 2 moles of $NI_3$ gives 3 moles of $I_2$

So, $9.069\times 10^{-3}$ moles of $NI_3$ gives $\frac{3}{2}\times (9.069\times 10^{-3})=0.01360$ moles of $I_2$

Therefore, the number of moles of $I_2$ form will be, 0.01360 moles