The percentage by mass of Na2CO3 in the sample is 48%.
The equation of the reaction of Na2CO3 with HCl;
Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)
Since the HCl is in excess, the excess is back titrated with NaOH as follows;
NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)
Number of moles of HCl added = 0.100 M × 50/1000 L = 0.005 moles
Number of moles of NaOH added = 5.6/1000 × 0.100 M = 0.00056 moles
Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.
Amount of HCl that reacted with Na2CO3 = 0.005 moles - 0.00056 moles = 0.0044 moles
Now;
1 mole of Na2CO3 reacts with 2 moles of HCl
x moles of Na2CO3 reacts with 0.0044 moles of HCl
x = 1 mole × 0.0044 moles / 2 moles
x = 0.0022 moles
Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g
Percentage of Na2CO3 in the sample = 0.24 g/ 0.500-g × 100/1 = 48%
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