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Serggg [28]
3 years ago
15

450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS

: 6.0*10^2g K2SO4) I need the working and there are a couple more of these if someone's tryna help my dumb self.
Chemistry
1 answer:
ExtremeBDS [4]3 years ago
8 0

Answer:

1597.959 g  

Explanation:

Given Data:

Amount of Cr₂(SO₄)₃ = 450 g

Amount of potassium phosphate K₃PO₄ = in Excess

grams of potassium sulfate K₂SO₄= ?

Solution

The Reaction will be

                 Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄

Information that we have from reaction

                Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄

                    1 mol          2 mol                   3 mol

we come to know from the above reaction that

1 mole of chromium(iii) sulfate (Cr₂(SO₄)₃) react with 2 mole of potassium phosphate (K₃PO₄) to produce 3 mole of K₂SO₄

We also know that

molar mass of Cr₂(SO₄)₃ = 147 g/mol

molar mass of K₃PO₄ = 212 g/mol

molar mass of K₂SO₄ = 174 g/mol

if we represent mole in grams then

      Cr₂(SO₄)₃             +       2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄

       1 mol (147 g/mol)         2 mol  (212 g/mol)      3 mol  (174g/mol)

So, Now we have the following details

          Cr₂(SO₄)₃  +  2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄

              147 g         424 g                           522 g

So,

we come to know that 147 g of Cr₂(SO₄)₃ combine with 424 g of 2K₃PO₄ produce  522 g of K₂SO₄

So now we calculate that how many grams of potassium sulfate will be produced

Apply unity formula

              147 g of  Cr₂(SO₄)₃  ≅ 522 g of K₂SO₄

              450 g of  Cr₂(SO₄)₃  ≅ ? g of K₂SO₄

by doing cross multiplication

g of K₂SO₄ =522 g x 450 g / 147 g

g of K₂SO₄ =  1597.959 g

So the write answer is  1597.959 g  

***Note: By calculation it is obvious that the correct answer is  1597.959 g  

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Step 1

The osmotic pressure is calculated as follows:

\begin{gathered} \pi\text{ = C x R x T} \\ C\text{ = molarity = }\frac{moles\text{ of solute}}{Volume\text{ of solution \lparen L\rparen}} \\ R\text{ = 0.082 }\frac{atmxL}{mol\text{ x K}} \end{gathered}

-------------

Step 2

<em>Information provided:</em>

The mass of solute = 13.6 g

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-------------

Step 3

Procedure:

1 L = 1000 mL => Volume = 251 mL x (1 L/1000 mL) = 0.251 L

---

C = moles of solute/volume of solution (L)

C = mass of solute/(molar mass x Volume (L))

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---

π = C x R x T

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