Calculation from Chemical Equations

Example 9.3.1:

Sodium combines with oxygen as follows:

\( \scriptsize 4Na_{(s)} \: + \: O_{2(g)} \: \rightarrow \: 2Na_2O_{(s)} \)

What is the mass of oxygen needed to burn 4.6 g of sodium?

(Na = 23, O = 16)

Solution:

From the Equation:

4 moles of Na reacts with 1 mole of O₂

Mass of 4 moles of Na = 4 × 23 = 92 g

Mass of 1 mole of O₂ = 16 × 2 = 32 g

92 g of Na reacts with 32 g of O₂

Let the mass of oxygen needed to burn 4.6 g of sodium be equal to x

4.6 g of Na reacts with x ⇒ \( \frac{92}{32} = \frac{4.6}{x} \\ \scriptsize x = \normalsize \frac{4.6 \: \times \: 32}{92 \: \times \: 1}\)

x = 1.6 g of O₂

Example 9.3.2:

a. What volume of oxygen, measured at STP is needed for the complete combustion of 20.0 g of hydrogen?

b. What is the mass of water produced in the reaction above?

(Molar volume at STP = 22.4 dm³, H = 1, O = 16)

Equation for the reaction:

\( \scriptsize 2H_{2(g)} \: + \: O_{2(g)} \: \rightarrow \: 2H_2O_{(l)} \)

Solution

a. If (2 × 2) of hydrogen requires (16 × 2) g of O₂

Therefore, 20 g of hydrogen will require:

\( \frac{20}{4} \: \times \: \frac{32}{1} \)

= 160 g of O₂

32 g of oxygen at STP occupies 22.4 dm³

Therefore 160 g of oxygen at STP will occupy

\( \frac{160}{32} \: \times \: \frac{22.4}{1} \\ \scriptsize = 112 \: dm^3 \: of \: oxygen \)

Or

Since we are required to find the volume of oxygen, molar mass of oxygen will not be used in the calculation.

If (2 × 2) g of hydrogen occupies 22.4 dm³ of oxygen at STP

Therefore, 20 g of hydrogen will occupy

\( \frac{20}{4} \: \times \: \frac{22.4}{1} \\ \scriptsize = 112 \: dm^3 \: of \: oxygen \)

b. \( \scriptsize 2H_{2(g)} \: + O_{2(g)} \: \rightarrow \: 2H_2O_{(l)} \)

If (2 × 2) g of hydrogen-induced 2(18) g of H₂O

Therefore, 20 g of hydrogen will produce:

\( \frac{20\:g}{4\:g} \: \times \: \frac{36\:g}{1} \\ \scriptsize = 180\:g \: of \: water \)