In the reaction $2 N a O H + H_{2} S O_{4} \to 2 H_{2} O + N a_{2} S O_{4}$ $2NaOH+H_2SO_4 -> 2H_2O+Na_2SO_4$ , 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?

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In the reaction $2 N a O H + H_{2} S O_{4} \to 2 H_{2} O + N a_{2} S O_{4}$ $2NaOH+H_2SO_4 -> 2H_2O+Na_2SO_4$ , 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?

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Answer 1 · 2016-03-15T02:01:58

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:

$M r_{N a O H} = 40 \frac{g}{m o l}$ $Mr_(NaOH)=40 g/(mol)$
$M r_{N a_{2} S O_{4}} = 142 \frac{g}{m o l}$ $Mr_(Na_2SO_4)=142 g/(mol)$

The analogy of the moles will be held constant:

$\frac{n_{N a O H}}{n_{N a_{2} S O_{4}}} = \frac{2}{1}$ $n_(NaOH)/n_(Na_2SO_4)=2/1$

$\frac{n_{N a O H}}{n_{N a_{2} S O_{4}}} = 2$ $n_(NaOH)/n_(Na_2SO_4)=2$

For each one, substitute:

$n = \frac{m}{M r}$ $n=m/(Mr)$

Therefore:

$\frac{n_{N a O H}}{n_{N a_{2} S O_{4}}} = 2$ $n_(NaOH)/n_(Na_2SO_4)=2$

$\frac{\frac{m_{N a O H}}{M r_{N a O H}}}{\frac{m_{N a_{2} S O_{4}}}{M r_{N a_{2} S O_{4}}}} = 2$ $(m_(NaOH)/(Mr_(NaOH)))/(m_(Na_2SO_4)/(Mr_(Na_2SO_4)))=2$

$\frac{\frac{200}{40}}{\frac{x}{142}} = 2$ $(200/40)/(x/142)=2$

$\frac{200 \cdot 142}{40 x} = 2$ $(200*142)/(40x)=2$

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

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

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

Answer 2 · 2016-03-15T02:31:38

Through conversion method using mole ratio and formula masses the answer is $355$ $355$ g of $N a_{2} S O_{4}$ $Na_2SO_4$

Explanation:

Equation is already balanced, therefore what you need to find are the formula masses of the involved compounds, $N a O H$ $NaOH$ and $N a_{2} S O_{4}$ $Na_2SO_4$ ;
Once known, start the calculation by converting $200$ $200$ g $N a O H$ $NaOH$ to mole $N a O H$ $NaOH$ by multiplying it with the ratio of the formula mass of $N a O H$ $NaOH$ ;
The result from the above calculation, will then be multiplied by the mole ratio of $N a_{2} S O_{4}$ $Na_2SO_4$ and $N a O H$ $NaOH$ , which is $\frac{1 m o l N a_{2} S O_{4}}{2 m o l N a O H}$ $(1 mol Na_2SO_4)/(2 mol NaOH)$ ;
Since, we are asked to find the mass of $N a_{2} S O_{4}$ $Na_2SO_4$ formed in this reaction, we need to multiply the answer of $s t e p 3$ $step 3$ to the ratio of the formula mass of $N a_{2} S O_{4}$ $Na_2SO_4$ .
Per calculation, the answer in mass is $355$ $355$ grams of $N a_{2} S O_{4}$ $Na_2SO_4$ .

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In the reaction $2 N a O H + H_{2} S O_{4} \to 2 H_{2} O + N a_{2} S O_{4}$ $2NaOH+H_2SO_4 -> 2H_2O+Na_2SO_4$ , 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?

2 Answers

Explanation:

Explanation:

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Explanation:

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In the reaction $2 N a O H + H_{2} S O_{4} \to 2 H_{2} O + N a_{2} S O_{4}$ $2NaOH+H_2SO_4 -> 2H_2O+Na_2SO_4$ , 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?