Respuesta :

the balanced chemical equation will b
N2H4 + 2 H2O2-> N2 + 4H2O
 so first of all we have to find the limiting reactant.
so for this 

9.24 g H2O2 ( 1 mol / 34.02 g mass of H2O2 )
 = 0.27 mol H2O2
Now
6.56  g of  N2H4 ( 1mol / 32.06) = 0.20 mol N2H4

so from the above reaction we found 1:2 ratio of the reactants
 so the limiting reactant is hydrogen peroxide. by using this we will find the amount of N2 produced.
so that will be

0.27 mol H2O2 ( 1 mol N2 / 2 mol H2O2 ) ( 14.01 g N2 / 1 mol N2)
 =1.89 g N2
hope it helps

Answer: The mass of nitrogen gas produces is 3.81 grams.

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

  • For hydrazine:

Given mass of hydrazine = 6.56 g

Molar mass of hydrazine = 32.0 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of hydrazine}=\frac{6.56g}{32.0g/mol}=0.205mol[/tex]

  • For hydrogen peroxide:

Given mass of hydrogen peroxide = 9.24 g

Molar mass of hydrogen peroxide = 34.0 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of hydrogen peroxide}=\frac{9.24g}{34.0g/mol}=0.272mol[/tex]

  • The chemical equation for the reaction of hydrazine and hydrogen peroxide follows:

[tex]N_2H_4+2H_2O_2\rightarrow N_2+4H_2O[/tex]

By Stoichiometry of the reaction:

2 moles of hydrogen peroxide reacts with 1 mole of hydrazine

So, 0.272 moles of hydrogen peroxide will react with = [tex]\frac{1}{2}\times 0.272=0.136mol[/tex] of hydrazine

As, given amount of hydrazine is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrogen peroxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrogen peroxide produces 1 mole of nitrogen gas

So, 0.272 moles of hydrogen peroxide will produce = [tex]\frac{1}{2}\times 0.272=0.136mol[/tex] of nitrogen gas

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

Molar mass of nitrogen gas = 28.0 g/mol

Moles of nitrogen gas = 0.136 moles

Putting values in equation 1, we get:

[tex]0.136mol=\frac{\text{Mass of nitrogen gas}}{28.0g/mol}\\\\\text{Mass of nitrogen gas}=(0.136mol\times 28.0g/mol)=3.81g[/tex]

Hence, the mass of nitrogen gas produces is 3.81 grams.

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