last question #2 the mass of reactants before the reaction, add the mass of sodi
ID: 712146 • Letter: L
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last question #2
the mass of reactants before the reaction, add the mass of sodium carbonate added to the mass of the flask and sulfuric acid. Number of Moles of Mass of Flask Erlenmeyer Before the Flask # Mass of Co2 Number of Moles Released (g) of cO2 Released Reaction (g) Na CO 111.650 112.650 114.650 116.650 118.650 120.650 122.650 0.0094349 0.0188697 0.0377394 0.0566091 0.0754788 0.0943485 0.01022494 0.01965462 0.03851392 0.05739604 0.06078164 0.06078164 0.06078164 10 D. 1132182 12 3. Create and save a graph of the number of moles of CO2 formed (v-axis) versus the number of moles of Na,COs added (x-axis) Click the graphing icon below to create your graph. the graph to identify the limiting reactant in each of the flasks. Use Conclusions Suppose you used excess sulfuric acid in all the flasks. How many moles of CO2 would be released in flask 7? The molar mass of sodium carbonate is 105.99 g/mol. 1. 2. Su ppose you were to perform a similar experiment using hydrochloric acid instead of sulfuric iven the reactant quantities in the table below, how many grams of carbon diox be formed in the reaction? The nm molar mass of carbon dioxide is 44.01 g/mol. olar mass of sodium carbonate is 105.99 g/mol and the mass of sodium carbonate volume of 4.00 M HCI solution 10.00 mL SUBMIT FOR GRADING SAVE AND FINISH LATER DiscardExplanation / Answer
2) Write the balanced chemical equation for the reaction between Na2CO3 and HCl as below.
Na2CO3 + 2 HCl ---------> 2 NaCl + CO2 + H2O
As per the stoichiometric equation,
1 mole Na2CO3 = 2 moles HCl = 1 mole CO2.
Mole(s) of Na2CO3 corresponding to 3.000 g Na2CO3 = (3.00 g)/(105.99 g/mol) = 0.02830 mole.
Mole(s) of HCl corresponding to 10.00 mL of 4.00 M HCl = (10.00 mL)*(1 L/1000 mL)*(4.00 M) = 0.04 mole.
Determine the limiting reactant.
Na2CO3: (0.02830 mole Na2CO3)*(2 mole HCl/1 mole Na2CO3) = 0.05660 mole HCl.
HCl: (0.04 mole HCl)*(1 mole Na2CO3/2 mole HCl) = 0.02 mole Na2CO3.
Offcourse, we do not have 0.05660 mole HCl, but we do have more than 0.02 mole Na2CO3. Therefore, HCl is the limiting reactant and the yield of CO2 is determined by the limiting reactant.
CO2: (0.04 mole HCl)*(1 mole CO2/2 mole HCl) = 0.02 mole CO2.
The molar mass of CO2 is 44.01 g/mol; hence, the mass of CO2 formed in the reaction = (0.02 mole)*(44.01 g/mol) = 0.8802 g (ans).
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