A1.00 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.

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

Explanation:

    .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

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

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:

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

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

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

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

the product of the pressure and volume is also constant

2b2t

explanation:

2b2t