<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium
<u>Explanation:</u>
- To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of NaOH solution = 0.5000 M
Volume of solution = 0.03340 L
Putting values in equation 1, we get:
![0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol](https://tex.z-dn.net/?f=0.5000M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20NaOH%7D%7D%7B0.03340L%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20NaOH%7D%3D%280.5000mol%2FL%5Ctimes%200.03340L%29%3D0.01670mol)
- The chemical equation for the reaction of NaOH and sulfuric acid follows:
![2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O](https://tex.z-dn.net/?f=2NaOH%2BH_2SO_4%5Crightarrow%20Na_2SO_4%2BH_2O)
By Stoichiometry of the reaction:
2 moles of NaOH reacts with 1 mole of sulfuric acid
So, 0.01670 moles of NaOH will react with =
of sulfuric acid
Excess moles of sulfuric acid = 0.00835 moles
- Calculating the moles of sulfuric acid by using equation 1, we get:
Molarity of sulfuric acid solution = 0.5000 M
Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
![0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol](https://tex.z-dn.net/?f=0.5000M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DH_2SO_4%7D%7B0.1279L%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20%7DH_2SO_4%3D%280.5000mol%2FL%5Ctimes%200.1279L%29%3D0.06395mol)
Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles
- The chemical equation for the reaction of metal (forming
ion) and sulfuric acid follows:
![2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2](https://tex.z-dn.net/?f=2X%2B3H_2SO_4%5Crightarrow%20X_2%28SO_4%29_3%2B3H_2)
By Stoichiometry of the reaction:
3 moles of sulfuric acid reacts with 2 moles of metal
So, 0.0556 moles of sulfuric acid will react with =
of metal
- To calculate the molar mass of metal for given number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Mass of metal = 1.00 g
Moles of metal = 0.0371 moles
Putting values in above equation, we get:
![0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol](https://tex.z-dn.net/?f=0.0371mol%3D%5Cfrac%7B1.00g%7D%7B%5Ctext%7BMolar%20mass%20of%20metal%7D%7D%5C%5C%5C%5C%5Ctext%7BMolar%20mass%20of%20metal%7D%3D%5Cfrac%7B1.00g%7D%7B0.0371mol%7D%3D26.95g%2Fmol)
Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium