10. How any grams of N He are required to produce 120 grams of HNO3, assuming H2O2 is present in excess?
N2H4 + 7 H2O2 → 2 HNO3 + 8 H2O
H2O2
N2H4
32
HNO3
63
H2O
18
34
Molar Mass
(g/mol)

30.48g of N2H4

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

N2H4 + 7H2O2 → 2HNO3 + 8H2O

Step 2:

Determination of the mass of N2H4 that reacted and the mass HNO3 produced from the balanced equation. This is illustrated below:

Molar mass of N2H4 = (14x2) + (4x1) = 32g/mol

Mass of N2H4 from the balanced equation = 1 x 32 = 32g

Molar Mass of HNO3 = 1 + 14 + (16x3) = 63g/mol

Mass of HNO3 from the balanced equation = 2 x 63 = 126g

Summary:

From the balanced equation above,

32g of N2H4 reacted to produce 126g of HNO3.

Step 3:

Determination of the mass of N2H4 required to produce 120g of HNO3. This is illustrated below:

From the balanced equation above,

32g of N2H4 reacted to produce 126g of HNO3.

Therefore, Xg of N2H4 will react to produce 120g of HNO3 i.e

Xg of N2H4 = (32 x 120)/126

Xg of N2H4 = 30.48g

Therefore, 30.48g of N2H4 is required to produce 120g of HNO3.

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