Lesson 9, Topic 3
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# Calculation from Chemical Equations

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Example I

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 4.6g 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 x 23 = 92g

Mass of 1 mole of O2 = 16 x 2 = 32g

92g of Na reacts with 32g of O2

4.6g of Na reacts with $$\frac{4.6 \; \times \; 32}{92 \; \times \; 1}$$

= 1.6g of O2

Example II

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

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

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

Equation for the reaction

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

Solution

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

Therefore, 20g of hydrogen will require

$$\frac{20}{4} \: \times \: \frac{32}{1}$$

= 160g of O2

32g of oxygen at STP occupies 22.4dm3

Therefore 160g of oxygen at STP will occupy

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

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 occupy 22.4dm3 of oxygen at STP

Therefore, 20g of hydrogen will occupy

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

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

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

Therefore, 20g of hydrogen will produce

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

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