Question

calculate the mass of salt produced when 0.2 M, 100 ml of H2So4 is mixed with 0.5 m 100 ml of naoh solution​

Answer 1

Answer:

2.84 grams of sodium sulfate salt is produced.

Explanation:

$Molarity=\frac{n}{\text{Volume of the solution in L}}$

n = number of moles of compound

$H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

Moles of sulfuric acid :

$0.2 M=\frac{n}{100\times 0.001 L]$

n = 0.02 moles

Moles of sodium hydroxide:

$0.5 M=\frac{n'}{100\times 0.001 L]$

n' = 0.05 moles

According to reaction 1 mol of sulfuric acid reacts with 2 moles of sodium hydroxide.

The 0.02 moles of sulfuric acid will react with:

$\frac{2}{1}\times 0.02 mol=0.04 mol$ of sodium hydroxide.

Since 0.05 moles of sodium hydroxide are present. Hence an excess reagent. moles of salt will depend upon moles of sulfuric acid .

According to reaction 1 mol of sulfuric acid gives 1 mol of sodium sulfate :

Then 0.02 moles of sulfuric acid will give:

$\frac{1}{1}\times 0.02 mol=0.02 mol$of sodium hydroxide

Mass of 0.02 moles of sodium hydroxide;

$0.02 mol\times 142 g.mol=2.84 g$

2.84 grams of sodium sulfate salt is produced.

Answer 2

Please refer to the image given below.