Question

How many grams of i2 are needed to react with 30.1 g of n2h4?

Answer 1

The reaction is,

H2S + I2 --------------> 2 HI +S

Molar weight of H2S = 34 g per mol

Molar weight of HI =128 g per mol

Molar weight of I2 =254 g per mol

Moles of H2S in 49.2 g = 49.2 /34 mol = 1.447 mol

So according to stoichiometry of the reaction, number of I2 mols needed

= 1.447 mol

The mass of I2 needed = 1.447 mol x 254 g

Answer 2

Answer: The mass of iodine gas needed to react is 477.14 g

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of
`N_2H_4` = 30.1 g

Molar mass of
`N_2H_4` = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of }N_2H_4=(30.1g)/(32g/mol)=0.94mol`

The chemical equation for the reaction of
`N_2H_4` and iodine gas follows:

`2I_2+N_2H_4\rightarrow 4HI+N_2`

By Stoichiometry of the reaction:

1 mole of
`N_2H_4` reacts with 2 moles of iodine gas

So, 0.94 moles of
`N_2H_4` will react with =
`(2)/(1)* 0.94=1.88mol` of iodine gas

Now, calculating the mass of iodine gas by using equation 1, we get:

Molar mass of iodine gas = 253.80 g/mol

Moles of iodine gas = 1.88 moles

Putting values in equation 1, we get:

`1.88mol=\frac{\text{Mass of iodine gas}}{253.80g/mol}\\\\\text{Mass of iodine gas}=(1.88mol* 253.80g/mol)=477.14g`

Hence, the mass of iodine gas needed to react is 477.14 g