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.

Question

Chemistry

Miguel Wilcox

Yesterday, 03:54

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.

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Answers (2)

Know the Answer?

Taryn Heath · Answer 1

95.59%

Explanation:

Calculate the theoretical yield when 1.54 * 103 g of V2O5 react with 1.96 * 103 g of Ca.

First, we find the limiting reactant:

1.54 * 103 g · 1 mol/181.881 g = 8.47 mol V2O5

1.96 * 103 g · 1 mol/40.078 g = 48.9 mol

The number of moles of Ca to react completely with 8.47 mol V2O5 is 8.47 mol V2O5 · 5 mol Ca/1 mol V2O5 = 42.4 mol

So Ca is in excess and we should use the molar amount of V2O5 to determine the theoretical yield of V.

8.47 mol V2O5 · 2 mol V/1 mol V2O5 = 16.9 mol V

Percent Yield What is the percent yield if 825 g of V are obtained?

First, find out how many grams the theoretical yield of V represents: 16.9 mol · 50.941 g/mol = 863 g

Then just divide the actual yield (given) by the theoretical yield: 825g/863g = 0.9559 * 100% = 95.59%

Yamilet Shaw · Answer 2

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 %