Short answer
stoichiometry by loss of co2
lab results
record the following lab results in the table below.
erlenmeyer flask # mass after reaction (g)
1 111.200
2 111.785
3 112.955
4 114.124
5 115.975
6 117.975
7 119.975
data analysis
for each reaction, calculate, and record the following data. the molar mass of sodium carbonate is 105.99 g/mol and the molar mass of carbon dioxide is 44.01 g/mol. to determine the mass of reactants before the reaction, add the mass of sodium carbonate added to the mass of the flask and sulfuric acid.
erlenmeyer flask # mass of flask before the reaction (g) number of moles of na2co3 mass of co2released (g) number of moles of co2released
1 111.65 9.44*10^-3 .4152 9.44*10^-3
2 112.615 18.87*10^-3 .8305 18.87*10^-3
3 114.615 37.74*10^-3 1.661 37.74*10^-3
4 116.615 56.61*10^-3 2.491 56.61*10^-3
5 118.615 75.48*10^-3 3.322 75.48*10^-3
6 120.615 94.35*10^-3 4.152 94.35*10^-3
7 122.615 113.22*10^-3 4.983 113.22*10^-3
create and save a graph of the number of moles of co2formed (y-axis) versus the number of moles of na2co3added (x-axis). click the graphing icon below to create your graph. use the graph to identify the limiting reactant in each of the flasks.
limiting reactant- na2co3
conclusions
suppose you used excess sulfuric acid in all the flasks. how many moles of co2would be released in flask 7? the molar mass of sodium carbonate is 105.99 g/mol.
suppose you were to perform a similar experiment using hydrochloric acid instead of sulfuric acid. given the reactant quantities in the table below, how many grams of carbon dioxide will be formed in the reaction? the molar mass of sodium carbonate is 105.99 g/mol and the molar mass of carbon dioxide is 44.01 g/mol.
mass of sodium carbonate 3.000 g
volume of 4.00 m hcl solution 10.00 ml
3.00g/ 105.99 g/mol= 0.0283 mol na2co3
10ml * (1/1000ml) *4.00 m= 0.04 mol
(0.02830 mole na2co3)*(2 mole hcl/1 mole na2co3) = 0.05660 mole hcl
(0.04 mole hcl)*(1 mole na2co3/2 mole hcl) = 0.02 mole na2co3
(0.04 mol hcl)*(1 mol co2/ 2 mol hcl) = .02 mol co2
(0.02 mol)*(44.01 g/mol) = 0.8802 g co2
need with conclusion question 1 and did i do question 2 right?