Na\(_2\)CO\(_3\)(g) + 2HCl(aq) → 2NaCl(aq) + H\(_2\)O(l) + CO\(_2\)(g)
How many moles of sodium trioxocarbonate (IV) are there in a 25 cm\(^3\) solution which required 10cm\(^3\) of 0.05 mol dm\(^3\) hydrochloric acid solution to neutralize it?
1.000 mole
0.100 mole
0.010 mole`
0.111 mole
Explanation
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Discussions (21)
pls the answer is not in the option, u guys are calculating concentration instead of moles. if u solve well u will get 2.5×10*-4moles
The question needs to be corrected..
How many numbers of mole is different from the concentration...
How many numbers of mole of Na2CO3 is 1 From the equation, which correspond with option A.
and the concentration is 0.01mol/dm³ which correspond with option C
I think, nA/nB = CAVA/n.
where nA/nB is mole ratio of the reacting mole of HCl to Na2CO3 = 2/1. CA is concentration of HCl given as 0.05mol/dm3 . VA is volume of HCl given as 10cm3 = 0.01dm3. and n is the number of mole of Na2CO3 required to neutralize the HCl.
So, n = CAVA/(nA/nB) = 0.05 × 0.01/2 = 2.5×10^-4mol.
There number of mole of CaCO3 required to completely neutralize 10cm3 of 0.05mol/dm3 concentration of HCl is 2.5×10^-4 mol
I think, nA/nB = CAVA/n.
where nA/nB is mole ratio of the reacting mole of HCl to Na2CO3 = 2/1. CA is concentration of HCl given as 0.05mol/dm3 . VA is volume of HCl given as 10cm3 = 0.01dm3. and n is the number of mole of Na2CO3 required to neutralize the HCl.
So, n = CAVA/(nA/nB) = 0.05 × 0.01/2 = 2.5×10^4mol.
There number of mole of CaCO3 required to completely neutralize 10cm3 of 0.05mol/dm3 concentration of HCl is 2.5×10^4 mol
CAVA/CBVB=NA/NB
We are looking for CB(concentration of base)
=CAVA*NB/VB*NA
=10*0.05/25*2=0.01 ans c.
Don't understand you guys? Correct solving but wrong option?
So this is how Jamb kept frustrating us abi?
