Suppose that for the same 10.0 ml sample described in q3, the mass of the crucible with precipitate was 17.550 g, and the mass o
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<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:
- The chemical equation for the reaction of NaOH and sulfuric acid follows:
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:
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:
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:
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