Question

Question #ef574

Answer 1

41.6 g

Explanation:

At first you need to have the whole balanced equation:
`H_2SO_4 + 2NaOH -> Na_2SO_4 + 2H_2O`
It is a neutralisation reaction and those have regular mechanism.

First we need to know how many moles of `H_2SO_4` we need.
For this we can use cross-multiplication
From the reaction we've written before we can see that 2 moles of `NaOH` react with one mole of `H_2SO_4` (no number means 1 mole)

`1 H_2SO_4 + 2NaOH ->1 Na_2SO_4 + 2H_2O`

So we use the cross multiplication
2 moles `NaOH` ..................................1 mole `H_2SO_4`
0.850 moles `NaOH`............................x mole `H_2SO_4`

`2*x= 1*0.850//:2`
`x=0.425` moles

Now that we have number of moles, we just need the molar mass (equation: `n=m/M` , where n= number of moles, m = mass, M = molar mass)

`M=2+(4*16)+32=98 g//mol`
After rearranging the equation: `m=n*M`
`m=0.425*98`
`m=41.6g`

If anything is unclear, feel free to ask :)

Answer 2

`41.68"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_2SO_4+2NaOHrarr Na_2SO_4+2H_2O`.

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"mol"` of `NaOH`:

`(2 "mol "NaOH)/(1 "mol " H_2SO_4)=(0.85"mol " NaOH)/(x)`

Cross-multiplying:

`1 "mol " H_2SO_4*0.85"mol " NaOH=x*2 "mol "NaOH`

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

`x=0.425"mol " H_2SO_4`

The formula for moles is `n=m/M`, where `n` is the moles, `m` the mass, and `M` the molar mass.

Rearranging to solve for `m`, we get:

`m=nM`.

Since we have `0.425"mol"` of sulfuric acid, and the molar mass of the acid is `98.08"g"/"mol"`.

We can input our values into the rearranged equation:

`m=0.425cancel("mol")*98.08"g"/cancel("mol")`

`m=41.68"g"` of sulfuric acid reacts with `0.85` moles of sodium hydroxide to produce water and sodium sulfate.