Question

Zinc reacts with nitric acid to produce hydrogen gas and zinc nitrate, Zn(NO3)2. 10.00 grams of zinc react with 7.84 grams of nitric acid. How many grams of excess reactant remain?

Answer 1

Zn is the excess reactant and remain 5.94g

Step-by-step explanation:

The reaction of Zn with HNO3 is:

Zn + 2 HNO3 → H2 + Zn(NO3)2

1 mole of Zn reacts with 2 moles of HNO3

To solve this question we must convert the mass of each reactant to moles in order to find limiting reactant. Thus, we can kknow the moles of excess reactant that will not react:

Moles Zn -Molar mass: 65.38g/mol-:

10.00g * (1mol / 65.38g) = 0.1530 moles Zn

Moles HNO3 -Molar mass: 63.01g/mol-:

7.84g * (1mol / 63.01g) = 0.124 moles HNO3

For a complete reaction of 0.124 moles of HNO3 are required:

0.124 moles HNO3 * (1mol Zn / 2 mol HNO3) = 0.0622 moles Zn. As there are 0.1530 moles, Zn is the excess reactant and will remain:

0.1530 moles - 0.0622 moles = 0.0908 moles Zn

The mass is:

0.0908 moles Zn * (65.38g / mol) =

5.94g