# Chemistry pls clearly step step for me

2gm of mixture hydrated sodium carbonate and sodium bicarbonate was dissolved in water and made up to 250cc. 25cc of this solution was titrated using methyl orange as indicator and 22.5cc of 0.087N hcl were required for naturalization. calculate the percentage of sodium bicarbonate in the mixture. I do not still understand.
Please clearly step by step for me.
I understand 25/250 x 2 = 0.2g
na2c03 = 106 /2 22.5cc x 0.087N = 1.957g
nahc03 = 84
x/53 + y/84 = 1.957g this part I cannot calculate to get the right answer
How get the right answer 56.7%

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1. And I can't calculate to obtain that answer either. I don't think the problem is worded correctly. Here is why I think that.
If the sample is 56.7% NaHCO3 then it is 43.3% Na2CO3.
g Na2CO3 = 2 x 0.433 = 0.866g Na2CO3
g NaHCO3 = 2 x 0.567 = 1.134g NaHCO3
total = 1.134+0.866 = 2.000g.

We titrated 0.1 of that so we titrated
0.0866g Na2CO3 + 0.1134g NaHCO3 and that is 0.0866+0.1134 = 0.2000 g titrated. How many mL of 0.087N HCl SHOULD THIS TAKE?
mL x N x mew = grams or
mL = g/(N*mew)
For Na2CO3 mL = 0.0866/(0.087*0.053) = 18.78 mL to titrate the 0.0866g Na2CO3.

For NaHCO3 mL = 0.1134/(0.087*0.084) = 15.52 mL to titrate the 0.1134g NaHCO3
.
mL for Na2CO3 + mL for NaHCO3 = total mL = 18.78+15.52 = 34.3 mL and for the problem that should be 22.5 mL. It isn't; therefore, the problem is not possible to work as is. If the problem is changed so that the HCl used was 34.3 mL of 0.087N then it is worked as follows:
eqn 1 is X + Y = 2
eqn 2 is [0.1X/0.053]+ [0.1Y/0.084] = 34.3*0.087

From eqn 2 we have
1.8867X + 1.19Y = 2.984
From eqn 1 if X+Y = 2 than X = 2-Y. Substitute that into eqn 2 as follows.
1.8867(2-Y) + 1.19Y = 2.984
3.7734 - 1.8867Y + 1.19Y = 2.984
-0.6976Y = -0.7893
Y = 1.133g NaHCO3
% = (1.133/2)*100 = 56.6% NaHCO3

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