Question

Nitric acid and zinc react to form zinc nitrate, ammonium nitrate, and water. How many atoms of zinc react with 1.49g HNO3? Calculate the number of grams of zinc that must react with an excess of HNO3 to form 29.1g NH4 NO3.

Answer 1

a) Molar mass of HNO3: 63 g/ mol
4 mol of Zn react with 10 mol of HNO3. So:
4 x 6.0x10^23 atoms of Zn ---------- 630 g of HNO3
x -------------------------------- 1.35 g
x= 5.1x10^21 atoms of Zn.

b)
Molar mass of Zn: 65.37 g/mol
Molar mass of NH4NO3: 80 g/mol

4 mol of Zn form 1 mol of NH4NO3. So:
4 x 65.37g of Zn ----------- 80 g of NH4NO3
m ----------------------- 27.5 g
m= 89.9 g of Zn.

Answer 2

For 1: The number of atoms of zinc are
`5.684* 10^(21)` atoms.

For 2: Mass of zinc reacted are 95.1279 grams.

Step-by-step explanation:

For the given chemical reaction, the equation follows:

`4Zn(s)+10HNO_3(aq.)\rightarrow 4Zn(NO_3)_2(aq.)+NH_4NO_3(aq.)+3H_2O(l)`

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

`\text{Number of moleS}=\frac{\text{Given mass}}{\text{Molar mass}}` ....(1)

• For 1:

For nitric acid:

Mass of nitric acid = 1.49 grams.

Molar mass of nitric acid = 63 g/mol

Putting values in equation 1, we get:

`\text{Moles of nitric acid}=(1.49g)/(63g/mol)\\\\\text{Moles of nitric acid}=0.0236moles`

By Stoichiometry of the reaction:

10 moles of nitric acid reacts with 4 moles of zinc metal.

So 0.0236 moles of nitric acid will react with =
`(4)/(10)* 0.0236=0.00944moles` of zinc metal

Now, according to Mole concept:

1 mole of an element contains
`6.022* 10^(23)` number of atoms.

So, 0.00944 moles of zinc will contain
`0.00944* 6.022* 10^(23)=5.684* 10^(21)` number of atoms.

Hence, the number of atoms of zinc are
`5.684* 10^(21)` atoms.

• For 2:

For ammonium nitrate:

Given mass of ammonium nitrate = 29.1 grams

Molar mass of ammonium nitrate = 80 g/mol

Putting values in equation 1, we get:

`\text{Moles of }NH_4NO_3=(29.1g)/(80g/mol)\\\\\text{Moles of }NH_4NO_3=0.36375 moles`

By Stoichiometry of the given reaction:

1 mole of ammonium nitrate is produced from 4 moles of zinc.

So, 0.36375 moles of ammonium nitrate will be produced from =
`(4)/(1)* 0.36375=1.455moles` of zinc.

Now, calculating the mass of zinc.

Moles of zinc = 1.455 moles

Molar mass of zinc = 65.38g/mol

Putting values in equation 1, we get:

`1.455mol=\frac{\text{Mass of zinc}}{65.38g/mol}\\\\\text{Mass of zinc}=95.1279g`

Hence, mass of zinc reacted are 95.1279 grams.