# 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?

Mar 15, 2016

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:

$M {r}_{N a O H} = 40 \frac{g}{m o l}$
$M {r}_{N {a}_{2} S {O}_{4}} = 142 \frac{g}{m o l}$

The analogy of the moles will be held constant:

${n}_{N a O H} / {n}_{N {a}_{2} S {O}_{4}} = \frac{2}{1}$

${n}_{N a O H} / {n}_{N {a}_{2} S {O}_{4}} = 2$

For each one, substitute:

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

Therefore:

${n}_{N a O H} / {n}_{N {a}_{2} S {O}_{4}} = 2$

$\frac{{m}_{N a O H} / \left(M {r}_{N a O H}\right)}{{m}_{N {a}_{2} S {O}_{4}} / \left(M {r}_{N {a}_{2} S {O}_{4}}\right)} = 2$

$\frac{\frac{200}{40}}{\frac{x}{142}} = 2$

$\frac{200 \cdot 142}{40 x} = 2$

$200 \cdot 142 = 2 \cdot 40 x$

$x = \frac{200 \cdot 142}{2 \cdot 40} = \frac{100 \cdot 142}{40} = 10 \cdot \frac{142}{4} = \frac{1420}{4} =$

$= \frac{710}{2} = 355 g r a m s$ (or just use a calculator)

Mar 15, 2016

Through conversion method using mole ratio and formula masses the answer is $355$ g of $N {a}_{2} S {O}_{4}$

#### Explanation:

1. Equation is already balanced, therefore what you need to find are the formula masses of the involved compounds, $N a O H$ and $N {a}_{2} S {O}_{4}$;
2. Once known, start the calculation by converting $200$ g $N a O H$ to mole $N a O H$ by multiplying it with the ratio of the formula mass of $N a O H$;
3. The result from the above calculation, will then be multiplied by the mole ratio of $N {a}_{2} S {O}_{4}$ and $N a O H$, which is $\frac{1 m o l N {a}_{2} S {O}_{4}}{2 m o l N a O H}$;
4. Since, we are asked to find the mass of $N {a}_{2} S {O}_{4}$ formed in this reaction, we need to multiply the answer of $s t e p 3$ to the ratio of the formula mass of $N {a}_{2} S {O}_{4}$.
5. Per calculation, the answer in mass is $355$ grams of $N {a}_{2} S {O}_{4}$.