Question #195789

A sample of an oxide of vanadium weighing 4.589 g was reduced with hydrogen gas to form water and another oxide of vanadium weighing 3.782 g. The second oxide was reduced further with hydrogen gas until only 2.573 g of vanadium metal remained. Write down balanced reaction equations for the two reduction reactions above.

Expert's answer

At first find the mass of vanadium and oxygen in the first oxide.

Let mass of element X in 4.589 g of first oxide is m(X) g.

Hence, mass of vanadium and oxygen are as follows:

m(V) = 2.573 g

m(O) = 4.589 g – 2.573 g = 2.016 g

To get number of moles n(X), divide mass of the element by molar mass of the same.

The molar mass of the vanadium is 50.94 g/mol.

Hence, the number of moles of vanadium is,

The molar mass of the oxygen is 16.0 g/mol.

Hence, the number of moles of oxygen is,

The number of moles of vanadium is 0.0505 mol and oxygen is 0.126 mol.

Now, divide each amount by the smallest amount 0.0505 mol.

Vanadium:

Oxygen:

As a compound always contains whole number of atoms, multiply each number by the smallest factor 2 that gives a whole number for each element to obtain the ratio 2:5.

Therefore, simplest formula of the first oxide is V_{2}O_{5}.

Find the mass of vanadium and oxygen in the second oxide.

Let mass of element X in 3.782 g of first oxide is m(X) g.

Hence, mass of vanadium and oxygen are as follows:

m(V) = 2.573 g

m(O) = 3.782 g – 2.573 g = 1.209 g

To get number of moles n(X), divide mass of the element by molar mass of the same.

The molar mass of the vanadium is 50.94 g/mol.

Hence, the number of moles of vanadium is,

The molar mass of the oxygen is 16.0 g/mol.

Hence, the number of moles of oxygen is,

Now divide each amount by the smallest amount.

Vanadium:

Oxygen:

As a compound always contains whole number of atoms, multiply each number by the smallest factor 2 that gives a whole number for each element to obtain the ratio 2:3.

Therefore, simplest formula of the second oxide is V_{2}O_{3}.

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