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Question: How much \(KN{{O}_{3}}\) must be heated to give as much \(O\), as required to burn 16 g of \(S\) int...

How much KNO3KN{{O}_{3}} must be heated to give as much OO, as required to burn 16 g of SS into SO2S{{O}_{2}}?
[N=14,O=16,K=39,S=32][N=14,O=16,K=39,S=32]
A. 100 g
B. 101 g
C. 91 g
D. 99.1 g

Explanation

Solution

KNO3KN{{O}_{3}} is known by the name potassium nitrate which is a chemical compound and known as an ionic salt of potassium ions represented by the symbol K+{{K}^{+}}and nitrate ions NO3N{{O}_{3}}^{-}. It generally occurs in nature as a mineral.

Complete answer:
A mole of a substance or particle can be defined as containing exactly 6.02214076×10236.02214076 \times {10^{23}} particles which may be atoms, molecules or ions where 6.02214076×10236.02214076 \times {10^{23}} is known as the Avogadro’s number.
S+O2SO2S+{{O}_{2}}\to S{{O}_{2}}
Moles of sulfur can be calculated by the formula
Given mass Molar mass\dfrac{Given\ \text{mass}\ }{Molar\ \text{mass}}; Molar mass is 32 and given is 16
Moles of sulphur =1632=0.5moleMoles\ of\ \text{sulphur =}\dfrac{16}{32}=0.5mole
This suggests that 0.5 mol of sulfur requires 0.5 mol of oxygen.
Now we know that sulphur burns in 0.5 mol oxygen.
Then the thermal decomposition of Potassium nitrate is given by the following reaction:
2KNO32KNO2+O22KN{{O}_{3}}\to 2KN{{O}_{2}}+{{O}_{2}}
This suggests that 1 mol of oxygen​ is produced by 2 mol of KNO3KN{{O}_{3}}
We can also say that 0.5mol of oxygen​ will be produced by 1 mol of KNO3KN{{O}_{3}}
Thus, 1 mole of KNO3KN{{O}_{3}}​ is required to produce the given amount of oxygen.
1 mol of KNO3KN{{O}_{3}}​= molar mass of KNO3KN{{O}_{3}}
Molar mass of KNO3KN{{O}_{3}}can be calculated by K+2×N+3×O=39+14×2+16×3=101K+2\times N+3\times O=39+14\times 2+16\times 3=101

Hence, option B is correct.

Note:
Mole is generally represented by the symbol mol. It is generally described as the unit of measurement for the amount of substance in SI where SI stands for International System of units. It is defined on the basis of Avogadro’s number.