Respuesta :
Answer:
For 1: The correct answer is 407.97 grams.
For 2: The correct answer is 76.72 grams.
Explanation:
- For 1:
We are given the molarity of HCl, to find the moles of HCl, we use the formula:
[tex]Molarity=\frac{\text{Moles}{\text{Volume}}[/tex]
We are given:
Molarity = 1.6 M
Volume = 7.8 L
Putting values in above equation, we get:
[tex]1.6mol/L=\frac{\text{Moles of HCl}}{7.8L}\\\\\text{Moles of HCl}=12.48mol[/tex]
For the given reaction:
[tex]Zn(s)+2HCl(aq.)\rightarrow ZnCl_2(aq.)+H_2(g)[/tex]
By Stoichiometry of the reaction:
2 moles of HCl reacts with 1 mole of zinc metal.
So, 12.48 moles of HCl react with = [tex]\frac{1}{2}\times 12.48=6.24mol[/tex] of Zinc metal.
To calculate the mass of zinc metal, we use the formula:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Molar mass of Zinc metal = 65.38 g/mol
[tex]6.24=\frac{\text{Mass of zinc metal}}{65.38g/mol}\\\\\text{Mass of zinc metal}=407.97g[/tex]
- For 2:
We are given that 298 grams of 83.1 % by mass of iron (II) nitrate solution are present.
So, mass of Iron (II) nitrate solution will be = [tex]\frac{83.1}{100}\times 298=247.638g[/tex]
Molar mass Iron (II) nitrate = 180 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }Fe(NO_3)_2=\frac{247.638g}{180g/mol}=1.37mol[/tex]
For the following reaction:
[tex]2Al(s)+3Fe(NO_3)_2(aq.)\rightarrow 3Fe(s)+2Al(NO_3)_3(aq.)[/tex]
By Stoichiometry of the reaction:
3 moles of [tex]Fe(NO_3)_2[/tex] produces 3 moles of iron metal.
So, 1.37 moles of [tex]Fe(NO_3)_2[/tex] will produce = [tex]\frac{3}{3}\times 1.37=1.37mol[/tex] of iron metal.
To calculate the mass of zinc metal, we equation 1:
[tex]1.37=\frac{\text{Mass of iron metal}}{56g/mol}\\\\\text{Mass of iron metal}=76.72g[/tex]
