Respuesta :
Answer: The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.
Explanation:
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 nickel:
Given mass of nickel = 14.8 g
Molar mass of nickel = 58.7 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol[/tex]
For the given chemical reaction:
[tex]3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)[/tex]
- For nickel (II) oxide:
By Stoichiometry of the reaction:
3 moles of nickel are produced from 3 moles of nickel (II) oxide
So, 0.252 moles of nickel will be produced from [tex]\frac{3}{3}\times 0.252=0.252mol[/tex] of nickel (II) oxide
Now, calculating the mass of nickel (II) oxide by using equation 1:
Molar mass of nickel (II) oxide = 74.7 g/mol
Moles of nickel (II) oxide = 0.252 moles
Putting values in equation 1, we get:
[tex]0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g[/tex]
- For aluminium:
By Stoichiometry of the reaction:
3 moles of nickel are produced from 2 moles of aluminium
So, 0.252 moles of nickel will be produced from [tex]\frac{2}{3}\times 0.252=0.168mol[/tex] of aluminium
Now, calculating the mass of aluminium by using equation 1:
Molar mass of aluminium = 27 g/mol
Moles of aluminium = 0.168 moles
Putting values in equation 1, we get:
[tex]0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g[/tex]
Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.