Answer:
2NaClO3 → 2NaCl + 3O2
It can be seen that 2 moles of NaClO3 produce 3 moles of O2.
Therefore, to find out how many moles of O2 are produced when 7.3 moles of NaClO3 decompose, we can set up a proportion:
2 mol NaClO3 / 3 mol O2 = 7.3 mol NaClO3 / x mol O2
Cross-multiplying, we get:
2 mol NaClO3 * x mol O2 = 3 mol O2 * 7.3 mol NaClO3
Simplifying:
x mol O2 = (3 mol O2 * 7.3 mol NaClO3) / 2 mol NaClO3
x mol O2 = 10.95 mol O2
Therefore, 10.95 moles of oxygen gas are produced when 7.3 moles of sodium chlorate decomposes.
Step-by-step explanation: