How do I calculate the percentage of Li2CO3 in the sample?

A sample of mixed carbonates, SrCO3 (147.63g/mol) and Li2CO3 (73.89 g/mol), weighed 0.5310 grams. If this sample is titrated with acid, it is found that 26.11ml of 0.3346 M HCl are required to reach the endpoint. What is the percentage of Li2CO3 in the sample? SrCO3 and Li2CO3 are the only components in the sample and carbonate reacts as a diprotic acid?How do I calculate the percentage of Li2CO3 in the sample?
you must start this problem with the information from the titration:



2H+ + (CO3)2- ----%26gt; H2CO3



therefore, the ratio of H+ to carbonate in the titration is 2:1. so the number of moles of carbonate in the sample = (0.3346moles/Liter *0.02611 Liters)/2 = 0.0.004568 moles Carbonate



now we have to find the amount of each component



number of moles SrCO3 = x

Number of moles Li2CO3 (Because it is total moles minus moles of SrCO3) = 0.004568 - x



147.63(x) + 73.89(0.004568- x) = 0.5310



The equation above is using the known mass of the total sample and the known total moles of carbonate to create an expression that will solve for the proportions of the two compounds that have different molar masses to get the total mass of the sample.



x = 0.00262368 moles SrCO3

Therefore

0.004568 - 0.00262368 = 0.001944 moles Li2CO3



Mass L2CO3 = (0.001944 moles * 73.89 g/mol) = 0.14367g





percent Li2CO3 (mass/mass total)*100

(0.14367g/0.5311g) *100 = 27.05% Li2CO3