Respuesta :
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.
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:
For 1: The number of atoms of zinc are [tex]5.684\times 10^{21}[/tex] atoms.
For 2: Mass of zinc reacted are 95.1279 grams.
Explanation:
For the given chemical reaction, the equation follows:
[tex]4Zn(s)+10HNO_3(aq.)\rightarrow 4Zn(NO_3)_2(aq.)+NH_4NO_3(aq.)+3H_2O(l)[/tex]
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 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:
[tex]\text{Moles of nitric acid}=\frac{1.49g}{63g/mol}\\\\\text{Moles of nitric acid}=0.0236moles[/tex]
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 = [tex]\frac{4}{10}\times 0.0236=0.00944moles[/tex] of zinc metal
Now, according to Mole concept:
1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms.
So, 0.00944 moles of zinc will contain [tex]0.00944\times 6.022\times 10^{23}=5.684\times 10^{21}[/tex] number of atoms.
Hence, the number of atoms of zinc are [tex]5.684\times 10^{21}[/tex] 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:
[tex]\text{Moles of }NH_4NO_3=\frac{29.1g}{80g/mol}\\\\\text{Moles of }NH_4NO_3=0.36375 moles[/tex]
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 = [tex]\frac{4}{1}\times 0.36375=1.455moles[/tex] 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:
[tex]1.455mol=\frac{\text{Mass of zinc}}{65.38g/mol}\\\\\text{Mass of zinc}=95.1279g[/tex]
Hence, mass of zinc reacted are 95.1279 grams.
