answer:
molarity = moles / volume of solution in L
So,
Moles of HCl reacted = 0.1025 * 0.03563 = 3.65*10-3 moles
Assume 'x' g of Na2CO3 is present and 'y' g of NaHCO3 in the 25mL aliquot.
So,
moles of Na2CO3 = x/106
moles of NaHCO3 = y/84
1 mole Na2CO3 requires 2 moles HCl, and 1 mole NaHCO3 requires 1 moles HCl
So,
2*(x/106) + 1*(y/84) = 3.65*10-3
Also, since original volume was 250 mL out of which this aliquot was taken, so actual masses are 10*x and 10*y.
Thus,
10*x + 10*y = 2.3684
Solve the above two equations to get the values of x and y
Assuming you weighed out a 2.3684g sample of the unknown, and it took 35.63 mL of...
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