Question

In the reaction `2NaOH+H_2SO_4 -> 2H_2O+Na_2SO_4`, how many grams of sodium sulfate will be formed if you start with 200.0 grams of sodium hydroxide and you have an excess of sulfuric acid?

Answer 1

Use the mole analogy of the reactant with the product. and substitute with the mole definition.

Answer is 355 grams.

Explanation:

Given the molecular weights:

`Mr_(NaOH)=40 g/(mol)`
`Mr_(Na_2SO_4)=142 g/(mol)`

The analogy of the moles will be held constant:

`n_(NaOH)/n_(Na_2SO_4)=2/1`

`n_(NaOH)/n_(Na_2SO_4)=2`

For each one, substitute:

`n=m/(Mr)`

Therefore:

`n_(NaOH)/n_(Na_2SO_4)=2`

`(m_(NaOH)/(Mr_(NaOH)))/(m_(Na_2SO_4)/(Mr_(Na_2SO_4)))=2`

`(200/40)/(x/142)=2`

`(200*142)/(40x)=2`

`200*142=2*40x`

`x=(200*142)/(2*40)=(100*142)/40=10*142/4=1420/4=`

`=710/2=355grams` (or just use a calculator)

Answer 2

Through conversion method using mole ratio and formula masses the answer is `355` g of `Na_2SO_4`

Explanation:

1. Equation is already balanced, therefore what you need to find are the formula masses of the involved compounds, `NaOH` and `Na_2SO_4`;
2. Once known, start the calculation by converting `200` g `NaOH` to mole `NaOH` by multiplying it with the ratio of the formula mass of `NaOH`;
3. The result from the above calculation, will then be multiplied by the mole ratio of `Na_2SO_4` and `NaOH`, which is `(1 mol Na_2SO_4)/(2 mol NaOH)`;
4. Since, we are asked to find the mass of `Na_2SO_4` formed in this reaction, we need to multiply the answer of `step 3` to the ratio of the formula mass of `Na_2SO_4`.
5. Per calculation, the answer in mass is `355` grams of `Na_2SO_4`.