Question

1. how many moles of oxygen are formed when 58.6 g of kno3 decomposes according to the following reaction? (molar mass of kno3 = 101.11 g/mol).

4 kno3(s) → 2 k2o(s) + 2 n2(g) + 5 o2(g)
a. 0.290 mol
b. 0.580 mol
c. 18.5 mol
d. 0.724 mol
e. 1.73 mol

Explanation:

Step1: Calculate moles of \(KNO_3\)

Use the formula \(n=\frac{m}{M}\), where \(m = 58.6g\) and \(M=101.11g/mol\).
\(n_{KNO_3}=\frac{58.6g}{101.11g/mol}\approx0.5796mol\)

Step2: Use mole ratio from balanced equation

From \(4KNO_3(s)\to2K_2O(s)+2N_2(g)+5O_2(g)\), the mole ratio of \(KNO_3\) to \(O_2\) is \(4:5\).
Let \(n_{O_2}\) be moles of \(O_2\). Then \(n_{O_2}=n_{KNO_3}\times\frac{5}{4}\)
Substitute \(n_{KNO_3} = 0.5796mol\)
\(n_{O_2}=0.5796mol\times\frac{5}{4}= 0.7245mol\approx0.724mol\)

Answer:

D. \(0.724\space mol\)