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?

0.009070.009400.0136 0.0212

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}=(3.58g)/(394.71g/mole)=9.069* 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* 10^(-3)` moles of
`NI_3` gives
`(3)/(2)* (9.069* 10^(-3))=0.01360` moles of
`I_2`

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