21.2 grams of N2H4 reacts with 29.2 grams of H202 to produce nitrogen gas and water. Which answer is correct?

a. H202 C.H20

b. N2 D. N2H4

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This reaction can be interpreted to be a form of single-replacement reaction. The equation below gives the reaction being described:

N2H4 + H2O2 --> N2 + H2O

This equation is not balanced yet, so let's go ahead and take care of that. You can make the equation into the following form:

**a**N2H4 + **b**H2O2 --> **c**N2 + **d**H2O

Notice that the coefficients with their atoms must be equivalent. This gives us the following equations based on the three elements in the equation:

Nitrogen: 2**a** = 2**c**

Hydrogen: 4**a** + 2**b** = 2**d**

Oxygen: 2**b** = **d**

Clearly, **a** = **c**, based on the nitrogen equation. However, the relationships bettween **a**, **b**, and **d** are a bit more complicated. To solve the relationship between these variables, we must substitute the last relation (**2b = d**) into the second relation for **d**:

4**a** + 2**b** = 2(2**b**)

Simplifying:

4**a** + 2**b** = 4**b**

Subtract 2**b** from both sides:

4**a** = 2**b**

Therefore, after dividing by 2, we establish the following relation between **a** and **b**:

2**a** = **b**

We can now express d in terms of a as well:

**d** = 2**b** = 2(2**a**) = 4**a**

Now, we can rewrite our equation in its balanced form (by letting **a** be 1):

N2H4 + **2**H2O2 --> N2 + **4**H2O

We now have our balanced equation, and you can use molar masses to determine the expected mass, yield, and moles of either product. However, the question (as it was written) did not make it clear what was being asked-for. The point in what I showed you above is that you can use that balanced equation, along with the periodic table, to answer any possible question they would likely have for that reaction.

I hope this starts you off on the right path!

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