Problem 6.139 A 1.268 g sample of a metal carbonate (MCO3) was treated with 100.
ID: 510256 • Letter: P
Question
Problem 6.139
A 1.268 g sample of a metal carbonate (MCO3) was treated with 100.00 mL of 0.1083 M sulfuric acid (H2SO4), yielding CO2 gas and an aqueous solution of the metal sulfate (MSO4). The solution was boiled to remove all the dissolved CO2 and was then titrated with 0.2145 MNaOH. A 32.41 mL volume of NaOH was required to neutralize the excess H2SO4.
Part A
What is the identity of the metal M?
Express your answer as a chemical formula.
SubmitMy AnswersGive Up
Part B
How many liters of CO2 gas were produced if the density of CO2 is 1.799 g/L?
Express your answer to foursignificant figures and include the appropriate units.
SubmitMy AnswersGive Up
Problem 6.139
A 1.268 g sample of a metal carbonate (MCO3) was treated with 100.00 mL of 0.1083 M sulfuric acid (H2SO4), yielding CO2 gas and an aqueous solution of the metal sulfate (MSO4). The solution was boiled to remove all the dissolved CO2 and was then titrated with 0.2145 MNaOH. A 32.41 mL volume of NaOH was required to neutralize the excess H2SO4.
Part A
What is the identity of the metal M?
Express your answer as a chemical formula.
SubmitMy AnswersGive Up
Part B
How many liters of CO2 gas were produced if the density of CO2 is 1.799 g/L?
Express your answer to foursignificant figures and include the appropriate units.
VCO2 =SubmitMy AnswersGive Up
Explanation / Answer
A 1.268 g sample of a metal carbonate (MCO3) was treated with 100.00 mL of 0.1083 M sulfuric acid (H2SO4), yielding CO2 gas and an aqueous solution of the metal sulfate (MSO4). The solution was boiled to remove all the dissolved CO2 and was then titrated with 0.2145 MNaOH. A 32.41 mL volume of NaOH was required to neutralize the excess H2SO4.
m = 1.268 g of MCO3
mol of acid = MV = 100*0.1083 = 10.83 mmol of H2SO4
mol of base = MV = 32.41*0.2145 = 6.95194 mmol of NaOH
excess H2SO4 = 6.95194/2 = 3.47597 mmol of H2SO4 were used in excess
so
mmol of H2SO4 actually used = 10.83 - 3.47597 = 7.35403 mmol of H2SO4
note that
1 mmol of H2SO4 = 1 mmol of MCO3
so
mmol of MCO3 = mmol of H2SO4 = 7.35403 mmol
so
mol = 7.35403 mol of MCO3
MW = mass/mol = 1.268/(7.35403*10^-3) = 172.422 g/mol
MW of CO3-2 = 60
MWof M = 172-60 = 112 G/mol
the metal is Cadmium
so
CdSO4
mass of CO2 produce =
mol of CdCO3 --> 1.268/(112+60) = 0.007372 moles of CO2
mass = mol*MW = 0.007372*44 = 0.324368 g of CO2
D = mass/V
V = mass/D = (0.324368)/(1.799) = 0.1803046137 liters, 180.3 mL