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garri49 [273]
3 years ago
7

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

Chemistry
2 answers:
Nina [5.8K]3 years ago
8 0

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

Step2247 [10]3 years ago
8 0

<u>Answer:</u> The mass of iodine gas needed to react is 477.14 g

<u>Explanation:</u>

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

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