Given the following reaction: H2SO4+2LiOH=Li2SO4+2H20, what mass of water is produced from 19 g of sulfuric acid?

Chemistry

1 answer:

goldfiish [28.3K]2 years ago

5 0

Hi,

To solve the question, first of all we will find out the no. of moles of H2SO4 in 19 g of sulfuric acid.

As we know:

No . of moles = Mass/ Molar mass

No. of moles= 19 g/98.08 g

No. of moles= 0.1937

Now we know the no of moles of H2SO4 that will react with 2LiOH. We also know the molar equivalence of H2SO4 , and 2LiOH that will react.

So, the water that will be produced will be 2H2O and 1 Li2SO4 when H2SO4 that will react with 2LiOH.

0.1937 x 2x 18.01

=6.977

=6.98

Therefore, approximately 6.98 grams of water will be produced from 19 g of sulfuric acid.

Hope it helps!

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See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

$M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol$

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

So you can solve for the moles of the solute:

$n_{solute}=M*V_{solution}$

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

$n_{solute}=1mol/L*1.5L=1.5mol$