how many ml of 0.1M HCL are required to rct comp with 1gm mixture of Na2CO3 and NaHCO3 containing equimolar solutions of both??
PLZ explain
106moles of Na2CO3 = x = moles of NaHCO3
==> y/106 = x = (1-y)/84
==> y = 0.56 gm = weight of Na2CO3 taken and weight of NaHCO3 taken = 0.44 gm
Now, 1 mol of Na2CO3 requires 2 moles HCl and 1 mol NaHCO3 requires 1 mol of HCl.
So, total moles of HCl required = 3x = 3*0.56/106
so, 0.1*z/1000 = 3*0.56/106
hence HCl required in ml = 3*5600/106 = 158.5 ml (approx)
1357Let the moles of Na2CO3 and that of NaHCO3 as n1 and n2 respectively
given equimolar solution
so M1=M2
implies n1=n2=n(let)
W1+W2=1
nM1+nM2=1
n=1/(M1+M2)=1/190
n mole of Na2CO3 requires 2n moles of HCl
n mole of NaHCO3 requires n moles of HCl
moles of HCl required=3n=3/190
let the volume be V
then MV=3n
V=30/190 litres=3000/19=158 ml