समानार्थी शब्द
रहस्य
Answers
Answer:
To Find
How many moles of potassium chlorate is to be heated to produce 11.2 litre oxygen
Solution
We know that
At NTP (Normal Temp. & Pressure)
22.4 L of oxygen = 1 mole
Given that
11.2 L of oxygen
Mathematically
\sf 22.4 \ L \longrightarrow 1 \ mole22.4 L⟶1 mole
\sf 11.2 \ L \longrightarrow "x" \ moles11.2 L⟶"x" moles
Calculating
Moles of oxygen (x)
By cross multiplication
We get
\sf \longrightarrow x= \dfrac{11.2}{22.4}=\dfrac{1}{2} \ moles⟶x=
22.4
11.2
=
2
1
moles
Therefore
11.2 L of oxygen = 0.5 moles
Chemical equation
\sf 2KClO_{3} \xrightarrow{\Delta} 2KCl+3O_{2}2KClO
3
Δ
2KCl+3O
2
From the chemical equation
We can understand that
2 moles of potassium chlorate on heating gives 3 moles of oxygen
According to the question
We are asked to find how many moles of potassium chlorate is to be heated to produce 11.2 litre oxygen
We found out that
11.2 L of oxygen = 0.5 mole
Mathematically
\sf 3 \ moles \ of \ O_{2} \longrightarrow 2 \ moles \ of \ KClO_{3}3 moles of O
2
⟶2 moles of KClO
3
\sf 0.5 \ moles \ of \ O_{2} \longrightarrow "y" \ moles \ of \ KClO_{3}0.5 moles of O
2
⟶"y" moles of KClO
3
Calculating
Moles of potassium chlorate (y)
By cross multiplication
We get
\sf \longrightarrow y= 0.5 \times \dfrac{2}{3} = \dfrac{1}{3} \ moles⟶y=0.5×
3
2
=
3
1
moles
Therefore
1/3 moles of potassium chlorate is to be heated to produce 11.2 L of oxygen
Hence
Option (b) is correct
Answer:
गोपनीयता, अंधकार, अस्पष्टता, छिपाव,