Answer to Question #195789 in General Chemistry for nathan

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. 


1
Expert's answer
2021-05-20T07:51:40-0400

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 \u2248 \\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 \u2248 \\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.


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