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Industrially, vanadium metal, which is used in steel alloys, can be obtained by reacting vanadium(V) oxide with calcium at high temperatures: 5Ca + V2O5 longrightarrow 5CaO + 2V In one process, 1.540 × 103 g of V2O5 react with 1.960 × 103 g Ca. Calculate the theoretical yield of V. Calculate the percent yield if 825.0 g of V are obtained.

Answer :

Answer:

The percent yield is 95.64 %

Explanation:

Step 1: Data given

Mass of V2O5 = 1540 grams

Mass of Ca = 1960 grams

actual yield of V = 825 grams

Molar mass V2O5 = 181.88 g/mol

Molar mass of Ca = 40.08 g/mol

Step 2: The balanced equation

5Ca + V2O5 → 5CaO + 2V

Step 3: Calculate moles

Moles = mass / molar mass

Moles V2O5 = 1540 grams / 181.88 g/mol

Moles V2O5 = 8.467 moles

Moles Ca = 1960 grams / 40.08 g/mol

Moles Ca = 48.90 moles

Step 4: Calculate limiting reactant

For 5 moles Ca we need 1 mol V2O5 to produce 2 moles V and 5moles CaO

V2O5 is the limiting reactant. It will completely be consumed (8.467 moles).

Ca is in excess. There will react 5*8.467 = 42.335 moles

There will remain 48.90 *42.335 = 6.565 moles

Step 5: Calculate moles V

For 5 moles Ca we need 1 mol V2O5 to produce 2 moles V and 5moles CaO

For 8.467 moles V2O5 we'll have 2*8.467 = 16.934 moles

Step 6: Calculate mass V

Mass V = moles V * molar mass V

Mass V = 16.934 moles * 50.94 g/mol

Mass V = 862.6 grams = theoretical yield

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (825.0 grams / 862.6 grams)*100%

% yield = 95.64 %

The percent yield is 95.64 %

pstnonsonjoku

The theoretical yield of V is 862.92 g and the percent yield is 95.6%.

The equation of the reaction is;

5Ca + V2O5 ------>  5CaO + 2V

Number of moles of Ca = 1.960 × 10^3 g /40 g/mol = 49 moles

Number of moles of V2O5 =  1.540 × 10^3 g/182 g/mol = 8.46 moles

Hence;

5 moles of Ca reacts with 1 mole of   V2O5

49 moles of Ca reacts with 49 moles �� 5 moles/ 1 mole

= 245 moles of V2O5

We can see that there is not enough V2O5 to react with Ca. So V2O5 is the limiting reactant.

1 mole of V2O5 yield 2 moles of V

8.46 moles of V2O5 yield 8.46 moles × 2 moles/1 mole

= 16.92 moles of V

Theoretical yield of V = 16.92 moles of V × 51 g/mol

= 862.92 g

% yield = actual yield/theoretical yield × 100/1

Actual yield =  825.0 g

% yield = 825.0 g/862.92 g × 100/1

% yield = 95.6%

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