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When vandium(V) oxide is reacted with hydrogen gas at elevated temperatures (155 degrees celsius and 1...

When vandium(V) oxide is reacted with hydrogen gas at elevated temperatures (155 degrees celsius and 1 atm), it produces vandium metal and water vapor in 92.4% yield. What mass of vandium(V) oxide and volume of hydrogen gas would be required to produce 11.3 grams of vandium under these conditions?

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Answer #1

mass of V = 11.3 g

moles of V = 11.3 / 50.94 = 0.2218 mol

% yield = (actual / theoretical) x 100

92.4 = (11.3 / theoretical) x 100

theoretical yield = 12.23 g

V2O5         +   5 H2    -----------> 2 V   + 5 H2O

181.88 g           10 g                       101.88 g

   ??                                                  12.23 g

mass of V2O5 = 12.23 x 181.88 / 101.88

mass of V2O5 = 21.8 g

mass of H2 formed = 1.2 g

moles of H2 = 0.600 mol

P V = n R T

1 x V = 0.600 x 0.0821 x (155 + 273)

V = 21.1 L

volume of H2 = 21.1 L

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