Answer to Question #195789 in General Chemistry for nathan

Question

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. 

Answer 1

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,

$n(V) = \frac{2.573}{50.94} = 0.0505 \;mol$

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

Hence, the number of moles of oxygen is,

$n(O) = \frac{2.016}{16.0} = 0.126 \;mol$

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:

$\frac{0.0505}{0.0505} = 1$

Oxygen:

$\frac{0.126}{0.0505} = 2.49 ≈ \frac{5}{2}$

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 V₂O₅.

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,

$n(V) = \frac{2.573}{50.94} = 0.0505 \;mol$

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

Hence, the number of moles of oxygen is,

$n(O) = \frac{1.209}{16.0}=0.07556 \;mol$

Now divide each amount by the smallest amount.

Vanadium:

$\frac{0.0505}{0.0505}=1$

Oxygen:

$\frac{0.07556}{0.0505}= 1.49 ≈ \frac{3}{2}$

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 V₂O₃.