The number of molecules of Carbon(iv)Oxide produced when 10.0g of CaCO\(_3\) is treated with 0.2dm\(^3\) of 1 Mole of HCL in the equation
CaCO\(_3\) + 2HCL ⇒ CaCl\(_2\) + H\(_2\) O + CO\(_2\) , is?
1.00 X 10\(^{23}\)
6.02 X 10\(^{23}\)
6.02 X 10\(^{22}\)
6.02 X 10\(^{24}\)
Explanation
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Discussions (15)
From the question we were asked to calculate the MOLE for carbon (iv) produced when 10.0g of CaCO3 is treated with O.2dm3 of 1 mole of HCL..
So now this equation is a bit tricky in my own way we are asked to calculate using CaCO3 to CO2 the just put that vol of HCL to confuse student's...(but you can actually use the both I.e HCL and CaCO3 to calculate for CO2 it will work)
Lemme solve it
CaCO3 + 2HCl ---- CaCL2 + H20 + CO2
Soln
Let me use CaCO3
n=mass/molar mass = 10/100 = 0.1mole of CaCo3
So, you compare their ratio..
So, 1 mole of CaCo3 =1 mole of CO2
0.1 mole of CaCO3 = x mole of CO2
So I crossed multiply
x = 0.1 mole of CO2
Remember we are asked to calculate the mole (molecules) of CO2.....
So, 1 mole of CaCO3 = 6.02×10²³ mole of of CO2
0.1 mole of CaCO3 = × mole of CO2
x= 6.02×10²³ ×0.1
6.02×10²³ × 1×10-¹
So our answer will be.
6.02×10²²
........
You can still calculate using volume I.e 0.2dm³ of HCL
Soln
So the equation is gon be HCL to CO2.......
The ratio there is 2:1
So, 2 mole of HCL = 1 mole of CO2
0.2 vol of HCL = x mole of CO2
0.2/2 = 0.1vol of CO2
So,
1 mole of HCL = 6.02×10²³
0.1 vol of CO2 = x
6.02 ×10²³ × 0.1
6.02× 10²³ × 1×10-¹
6.02×10²²
You see its the same thing....
But based on the real question we were asked to calculate for the mole of CO2
6.02 × 10²³ is Avogadro's number
it is constant
1 mole of any substance contains 6.02 × 10²³ (whether it is an atom, molecules or ions)
simply
1 mole of sub= 6.02 × 10²³
for those of u asking where the 6.0 × 10²³ come from 👍
