Question

The mass of 70%h2so4 required for neutralisation of one mole of naoh is:

Answer 1

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2NaOH + H2SO4 = Na2SO4 + H2O

Pure H2SO4 required for 1 mole of NaOH

= 1/2 mole = 49g.

70% H2SO4 required for 1 mole

= 49 x 100 /70.


= 70 g.


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Answer 2

Answer: The mass of $H_2SO_4$ required is 70 grams.

Explanation:

To calculate number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

The chemical equation for the neutralization of sulfuric acid and sodium hydroxide follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of sodium hydroxide is neutralized by 1 mole of sulfuric acid.

So, 1 mole of sodium hydroxide will be neutralized by = $\frac{1}{2}\times 1=0.5mol$ of sulfuric acid.

Calculating the mass of sulfuric acid required by using equation 1, we get:

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

$0.5mol=\frac{\text{Mass of sulfuric acid}}{98g/mol}\\\\\text{Mass of sulfuric acid}=49g/mol$

We are given:

(m/m) % of sulfuric acid = 70 %

This means that 70 grams of sulfuric acid is present in 100 g of solution.

So, 49 grams of sulfuric acid will be present in = $\frac{100}{70}\times 49=70g$

Hence, the mass of sulfuric acid required is 70 grams.