Question

A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After all the metal had reacted, the remaining acid required 0.03340 L of 0.5000 M NaOH solution for complete neutralization. Calculate the molar mass of the metal and identify the element.

Answer 1

Answer: The metal having molar mass equal to 26.95 g/mol is Aluminium

Step-by-step explanation:

• To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}` .....(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* 0.03340L)=0.01670mol`

• The chemical equation for the reaction of NaOH and sulfuric acid follows:

`2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_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 =
`(1)/(2)* 0.01670=0.00835mol` 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* 0.1279L)=0.06395mol`

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

• The chemical equation for the reaction of metal (forming
  `M^(3+)` ion) and sulfuric acid follows:

`2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_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 =
`(2)/(3)* 0.0556=0.0371mol` 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}}`

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}=(1.00g)/(0.0371mol)=26.95g/mol`

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium