# Question ef574

Feb 11, 2018

41.6 g

#### Explanation:

At first you need to have the whole balanced equation:
${H}_{2} S {O}_{4} + 2 N a O H \to N {a}_{2} S {O}_{4} + 2 {H}_{2} O$
It is a neutralisation reaction and those have regular mechanism.

First we need to know how many moles of ${H}_{2} S {O}_{4}$ we need.
For this we can use cross-multiplication
From the reaction we've written before we can see that 2 moles of $N a O H$ react with one mole of ${H}_{2} S {O}_{4}$ (no number means 1 mole)

$1 {H}_{2} S {O}_{4} + 2 N a O H \to 1 N {a}_{2} S {O}_{4} + 2 {H}_{2} O$

So we use the cross multiplication
2 moles $N a O H$ ..................................1 mole ${H}_{2} S {O}_{4}$
0.850 moles $N a O H$............................x mole ${H}_{2} S {O}_{4}$

$2 \cdot x = 1 \cdot 0.850 / : 2$
$x = 0.425$ moles

Now that we have number of moles, we just need the molar mass (equation: $n = \frac{m}{M}$ , where n= number of moles, m = mass, M = molar mass)

$M = 2 + \left(4 \cdot 16\right) + 32 = 98 g / m o l$
After rearranging the equation: $m = n \cdot M$
$m = 0.425 \cdot 98$
$m = 41.6 g$

If anything is unclear, feel free to ask :)

Feb 11, 2018

$41.68 \text{g}$ of sulfuric acid reacts with $0.85$ moles of sodium hydroxide to produce water and sodium sulfate.

#### Explanation:

First write down the reaction in full:

${H}_{2} S {O}_{4} + 2 N a O H \rightarrow N {a}_{2} S {O}_{4} + 2 {H}_{2} O$.

We won't need the equation, but it is clearer if you do so.

Notice how the moles of sulfuric acid and sodium hydroxide are in the ratio $2 : 1$. So if we have $0.85 \text{mol}$ of $N a O H$:

$\frac{2 \text{mol "NaOH)/(1 "mol " H_2SO_4)=(0.85"mol } N a O H}{x}$

Cross-multiplying:

$1 \text{mol " H_2SO_4*0.85"mol " NaOH=x*2 "mol } N a O H$

x=(1 "mol " H_2SO_4*0.85cancel("mol " NaOH))/(2 cancel("mol "NaOH))#

$x = 0.425 \text{mol } {H}_{2} S {O}_{4}$

The formula for moles is $n = \frac{m}{M}$, where $n$ is the moles, $m$ the mass, and $M$ the molar mass.

Rearranging to solve for $m$, we get:

$m = n M$.

Since we have $0.425 \text{mol}$ of sulfuric acid, and the molar mass of the acid is $98.08 \text{g"/"mol}$.

We can input our values into the rearranged equation:

$m = 0.425 \cancel{\text{mol")*98.08"g"/cancel("mol}}$

$m = 41.68 \text{g}$ of sulfuric acid reacts with $0.85$ moles of sodium hydroxide to produce water and sodium sulfate.