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

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:

$\text{Moles of }N_2H_4=\frac{30.1g}{32g/mol}=0.94mol$

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 = $\frac{2}{1}\times 0.94=1.88mol$ 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:

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

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