Given the balanced equation 2KC103+ 2KC1+302

The following information was recorded by a student team working to prepare nickel sulfate. Plan: Prepare NiSO4 by reacting NiO

mixas84 [53]

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 $NiSO_4.6H_2O$ :

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

3 0

3 years ago

If the sample has a total mass of 5.76 g and contains 1.79 g k, what are the percentages of kbr and ki in the sample by mass

Luden [163]

The percentages of KBr and KI in the sample by mass is 80.68 and 19.32 % respectively.

<h3>What is Molar Mass ?</h3>

Molar mass is defined as the mass contained in 1 mole of sample.

It is given that

the sample has a total mass of 5.76 g contains KBr and KI

and contains 1.79 gm of K is present

what is the percentage of KBr and KI in the sample

Molecular weight of K = 39

Molecular weight of Br = 79.9

Molecular weight of I = 126.9

In KBr the mass percentage of K is 39/(39+79.9) = 32.89%

In KI the mass percentage of K is 39/(39+126.9) = 23.5%

Let the mass of KBr present in the sample is x

K will be 0.3289 x

and let the mass of KI present be y

K will be 0.235y

x +y =5.76

0.3289x+0.235y = 1.79

0.0939y = 0.1045

y = 1.1125 gm

x = 5.76-1.1125

x = 4.6475 gm

% of KBr = (4.6475/5.76 )*100 = 80.68 %

% of KI = (1.1125/5.76) *100 = 19.32%

To know more about Molar Mass

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8 0

2 years ago

Prepare and name an alcohol by using the following reactants

Archy [21]

Answer:

Alcohol will form at more substituted carbon due to carbocation rearrangement so 2-hexanol

Explanation:

Well if it’s acid catalyzed

5 0

2 years ago

What is the correct formula for calcium sulfate dihydrate?

Murljashka [212]

Answer:

It is CaSO4.2H2O

Explanation:

Calcium has a valence of 2 because it has only two valency electrons.

Sulphate ( SO4)^2- is a radical with a valence value of 2.

When Calcium combines with sulphate in bonding,

Formular ==>  Ca2(SO4)2

But the 2 cancel out.

Overall formular==> CaSO4

For hydrated, ==> CaSO4.H2O

3 0

3 years ago

Give an example of a polar molecular compound that dissolves in water

VLD [36.1K]

Try sugar. I hope this helps!!!!!

7 0

3 years ago