Here are the Qs One of the reactions used in this lab actually occurs in two dis

Question

Here are the Qs

One of the reactions used in this lab actually occurs in two discrete steps, shown below. What is the overall net reaction that occurs? Na2CO3 (aq) + HCl (aq) rightarrow HCO3- (aq) + NaCl (aq) HCO3- (aq) + HCl (aq) rightarrow H2CO3 (aq) + NaCl (aq) Calculate the molar mass of CO 3 2- Determine the % (by mass) of carbonate in Na2CO3. In this experiment, you will conduct two different chemical reactions (see below). Some of the moles of HCl you add will react with sodium carbonate in the first reaction and the rest will react with sodium hydroxide in the second reaction. Na2CO3 (aq) + 2 HC1 (aq) rightarrow 2 NaCl (aq) + CO2 (g) + H2O (l) HC1 (aq) + NaOH (aq) rightarrow NaCl (aq) + H2O (1) If you use 0.502 mol HCl (total) and 0.252 mol NaOH in this experiment, answer the following: How many moles of the HCl react with NaOH? How many moles of the HCl react with Na2CO3? How many moles of Na2CO3 react with the HCl?

Explanation / Answer

1) over all net reaction => Na2CO3 +2HCl = H2CO3 + 2NaCl

2) molar mass of CO3 = 12+16*3 = 48+12 = 60 amu

% of CO3 in Na2CO3 = 60/(23+23+60) = 60/106 = 56.604%

3) a) no of moles of HCl used to react with NaOH = moles of NaOH = 0.252 moles

b) no of moles of HCl used to react with Na2CO3 = 0.502 -no of moles of HCl used to react with NaOH

= .250 moles

c) no of moles of Na2CO3 used to react with HCl = .250/2 = .125 moles

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