Question

The following reaction shows sodium hydroxide reacting with sulfuric acid.

Answer 1

its answer is 35.5 grams

Answer 2

Answer: The mass of $Na_2SO_4$ produced will be 17.8 grams.

Explanation:

To calculate the number of moles, we use the equation:

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

For Sodium hydroxide:

Given mass of sodium hydroxide = 10 g

Molar mass of $NaOH=[(1\times 22.989)+(1\times 15.999)+(1\times 1.008)]g/mol=39.996g/mol$

Putting values in above equation, we get:  

$\text{Moles of sodium hydroxide}=\frac{10g}{39.996g/mol}=0.25mol$

For the given chemical equation:

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

By Stoichiometry of the reaction:

2 moles of sodium hydroxide produces 1 mole of sodium sulfate.

So, 0.25 moles of sodium hydroxide will produce = $\frac{1}{2}\times 0.25=0.125moles$ of sodium sulfate.

Now, to calculate the mass of sodium sulfate, we use equation 1:

Moles of sodium sulfate = 0.125 moles

Molar mass of $Na_2SO_4=[(2\times 22.989)+(1\times 32.065)+(4\times 15.999)]g/mol=142.039g/mol$

Putting values in equation 1, we get:

$0.125mol=\frac{\text{Mass of }Na_2SO_4}{142.039g/mol}\\\\\text{Moles of }Na_2SO_4=17.8g$

Hence, the mass of $Na_2SO_4$ produced will be 17.8 grams.