Question

The following information was recorded by a student team working to prepare nickel sulfate. Plan: Prepare NiSO4 by reacting NiO with 6 M H2SO4 The expected product is NiSO4·6H2O. We will filter the crystals from the reaction solution to get the product. NiO + H2SO4 --> NiSO4 + H2O molar mass of NiO: 74.69 g/mol molar mass of NiSO4: 154.75 g/mol molar mass of NiSO4·6H2O: 262.85 g/mol mass of NiO used: 7.5 g volume of 6M H2SO4 used: 50 mL heated the solution for 30 minutes, cool and filtered the crystals mass of product: 17.4 g Two students calculated percent yield using this data. One student got 66.2% and another got 112%. Based on this information, which of the following statements are true.
a. Both percent yields are incorrect because the student used the wrong number of moles.
b. The percent yield of 112% was calculated using the molar mass for the hydrated product.
c. The percent yield of 66.2% was calculated using the molar mass for the anhydrous product.
d. The percent yield of 66.2% was calculated using the molar mass for the hydrated product.
e. The percent yield of 112% was calculated using the molar mass for the anhydrous product.

Answer 1

Answer:

The options e and d are correct.

Explanation:

Mass of NiO = 7.5 g

Moles of NiO = $\frac{7.5 g}{74.69 g/mol}=0.10 mol$

Moles of sulfuric acid = n

Volume of sulfuric acid ,V= 50 mL = 0.050 L

Molarity of sulfuric acid ,M = 6 mol/L

$n=M\time V=6mol/L\times 0.050 L =0.3 mol$

$NiO + H_2SO_4\rightarrow NiSO_4 + H_2O$

According to reaction, 1 mole of NiO reacts with 1 mole of sulfuric acid.

Then 0.10 moles of NiO reacts with :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of sulfuric acid.

As we can see that sulfuric acid is in excess amount, so the amount of the product will depend upon amount of NiO.

According to reaction, 1 mole of NiO gives with 1 mole of $NiSO_4$.

Then 0.10 moles of NiO wil give :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of  $NiSO_4$.

Molar mass of  $NiSO_4$ = 154.75 g/mol

Mass of 0.10 moles of $NiSO_4$:

= 154.75 g/mol × 0.10 mol = 15.475 g

Theoretical mass of $NiSO_4$ = 15.475 g

Experimental yield of $NiSO_4$ = 17.4 g

Percentage yield :

$\Yield=\frac{\text{Experimental mass}}{\text{Theoretical mass}}\times 100$

Percentage yield of $NiSO_4$:

$\Yield=\frac{17.4}{15.475 g}\times 100=112\%$

Moles of $NiSO_4.6H_2O$ = 262.85 g/mol × 0.10 mol = 26.285 g

Experimental yield of $NiSO_4.6H_2O$ = 17.4 g

Percentage yield of $NiSO_4.6H_2O$:

$\Yield=\frac{17.4}{26.285 g}\times 100=66.2\%$