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EXAMPLE 1.3.4

The decomposition of potassium nitrate is represented by the equation below:
[tex]\[
4 KNO_3(s) \rightarrow 2 K_2O(s) + 2 N_2(g) + 5 O_2(g)
\][/tex]

Calculate how many moles of [tex]\[ KNO_3 \][/tex] must be heated to produce 58 kg of oxygen.


Sagot :

Sure! Let's solve this problem step-by-step:

1. Understand the balanced chemical equation:

The balanced chemical equation for the decomposition of potassium nitrate ([tex]\( KNO_3 \)[/tex]) is:
[tex]\[ 4 \, \text{KNO}_3 (s) \rightarrow 2 \, \text{K}_2\text{O} (s) + 2 \, \text{N}_2 (g) + 5 \, \text{O}_2 (g) \][/tex]
This tells us that 4 moles of [tex]\( KNO_3 \)[/tex] produce 5 moles of [tex]\( O_2 \)[/tex].

2. Given information:

- We need to produce 58 kg of oxygen ([tex]\( O_2 \)[/tex]).
- The molar mass of oxygen ([tex]\( O_2 \)[/tex]) is 32 g/mol.

3. Conversion of mass of oxygen to grams:

Since the given mass is in kilograms, we need to convert it to grams:
[tex]\[ 58 \, \text{kg} = 58 \times 1000 = 58000 \, \text{grams} \][/tex]

4. Calculate moles of [tex]\( O_2 \)[/tex]:

To find out how many moles of [tex]\( O_2 \)[/tex] are produced, we use the molar mass of [tex]\( O_2 \)[/tex]:
[tex]\[ \text{Moles of } O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} \][/tex]
[tex]\[ \text{Moles of } O_2 = \frac{58000 \, \text{grams}}{32 \, \text{g/mol}} = 1812.5 \, \text{moles} \][/tex]

5. Determine the moles of [tex]\( KNO_3 \)[/tex] needed:

According to the balanced chemical equation, 4 moles of [tex]\( KNO_3 \)[/tex] produce 5 moles of [tex]\( O_2 \)[/tex]. We can set up a ratio to find the moles of [tex]\( KNO_3 \)[/tex] required:
[tex]\[ \frac{4 \, \text{moles of } KNO_3}{5 \, \text{moles of } O_2} = \frac{x \, \text{moles of } KNO_3}{1812.5 \, \text{moles of } O_2} \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 1812.5 \times \left(\frac{4}{5}\right) = 1450.0 \, \text{moles} \][/tex]

6. Conclusion:

The number of moles of [tex]\( KNO_3 \)[/tex] that must be heated to produce 58 kg of oxygen is:
[tex]\[ 1450.0 \, \text{moles} \][/tex]