Question

the molar mass of l2 is 253.80g/mol and the molar mass of nl3 is the 394.71g/mol. how many moles of l2 will form 38.58 g of ni3

Answer 1

To find the number of moles of I2 that will form 38.58 g of NI3, we use the molar mass of NI3 to convert the given mass to moles and then apply the stoichiometry of the reaction. We find that 0.0977 moles of NI3 would require a third of that amount on moles of I2, resulting in 0.03257 moles of I2.

Step-by-step explanation:

To calculate the number of moles of I2 that would form 38.58 g of NI3, we need to use the molar mass of NI3 and stoichiometry from the balanced chemical equation. The molar mass provided for NI3 is 394.71 g/mol.

First, we convert the mass of NI3 to moles using its molar mass:

moles of NI3 = mass of NI3 / molar mass of NI3 = 38.58 g / 394.71 g/mol

This gives us 0.0977 moles of NI3 (after rounding to four significant figures).

The balanced chemical equation for the formation of NI3 involves a 1:3 molar ratio between NI3 and I2:

3 NI -> NI3

Thus, the moles of I2 required would be one-third of the moles of NI3 produced. Therefore, we have:

moles of I2 = 0.0977 moles of NI3 / 3 = 0.03257 moles of I2 (to four significant figures)

Answer 2

The moles of I2 that will form 38.58g of Ni3 is 0.147 moles

calculation
write the equation for reaction for formation of Ni3 from I2 and N2

that is 3I2 + N2 = 2NI3

find the moles of Ni3 formed = mass/molar mass

= 38.58g/ 394.71 g/mol = 0.098 moles

by use of mole ratio between I2 to NI3 which is 3 :2 the moles of I2

= 0.098 x3/2 = 0.147 moles