thank you very much! Calculating the Percent Na_2CO_3 in Unknown B Suppose you d

Question

thank you very much! Calculating the Percent Na_2CO_3 in Unknown B Suppose you did several more standardization titrations and you found that the average molarity of your HCl was 0.09253 M. Then you used this standardized HCl to titrate 0.4567 grams of your Unknown B and it took 39.63 mL of your HCl to completely protonate the Na_2CO_3 in your 0.4567 g sample of Unknown B. How many moles of HCl are in the 39.63mL of your standardized HCl? How many moles of Na_2CO_3 were in your 0.4567 gram sample of Unknown B? How many grams of Na_2CO_3 were in your 0.4567 gram sample of Unknown B? What is the percent Na_2CO_3 in this sample of Unknown B?

Explanation / Answer

Balanced equation is,

2 HCl (aq.) + Na2CO3 (aq.) -----------> 2 NaCl (aq.) + CO2 (g) + H2O (l)

(1) Moles of HCl = Molarity * Volume in mL / 1000 = 0.09253 * 39.63 / 1000 = 0.003667 mol

(2) According to the balanced equation,

2 mol of HCl = 1 mol of Na2CO3

then, 0.003667 mol of HCl = 0.003667 * 1 / 2 = 0.001834 mol

So, moles of Na2CO3 in 0.4567 g. of unknown B = 0.001834 mol

(3) Mass of Na2CO3 = moles * molar mass of Na2CO3 = 0.001834 * 106 = 0.1944 g.

(4) % by mass of Na2CO3 = (0.1944 / 0.4567) * 100 = 42.57 %

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