Question

2NaOH + H2SO4 ------> 2 H2O + Na2SO4

Answer 1

$\huge\fbox{Answer ☘}$

It is given in the question that 200.0 grams of sodium hydroxide reacts with excess of sulphuric acid.

The given reaction is shown below

$\bold\blue{2NAOH + H _{2}SO _{4} \: - > 2H _{2} O+ NA _{2}SO_{4} }$

In this reaction, two mole of sodium hydroxide react with one mole of sulphuric acid to give two mole of water and one mole of sodium sulfate.

To determine the mass of sodium sulfate, first calculate the moles of sodium hydroxide as the value of mass is provided.

The molecular weight of NaOH is 40 g/mol.

The formula is shown below.

$\pink{n = \frac{m}{ M}}$

Where,

✫ n is the number of moles

✫ m is the mass

✫ M is the molecular weight

To calculate the number of moles, substitute the values in the above equation.

$\pink{n = \frac{200.0}{40g \: mol {}^{ - 1} } } \\ \\\pink{ ⇢n = 5}$

In the equation it is given that 2 mole of sodium hydroxide gives 1 mole of sodium sulphate.

So, to determine the actual mole of sodium sulfate divide the calculated moles of NaOH by 2.

$\pink{⇢ \frac{5}{2}} \\ \\ \pink{⇢ 2.5 \: mol}$

So 2.5 mol sodium sulfate is formed from 5 mol NaOH.

The molecular weight of sodium sulfate is 142.1 g/mol.

To calculate the mass of sodium sulfate, substitute the values in the formula.

$\pink{⇢2.5 = \frac{m}{142.1g \: mol {}^{ - 1} } } \\ \\\pink{ ⇢ m = 2.5 \times 142.1 }\\ \\ \pink{⇢ m = 355.25}$

Therefore, the mass of sodium sulfate formed when we start with 200.0 grams of sodium hydroxide and you have excess sulphuric acid is 355.25g.

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