urgent chemistry help
Hey!
A 0.4108g sample of calcium carbonate is added to a flask which contains 15.00cm³ of 2.00 mol dm^-3 hydrochloric acid.
CaCO3 +2HCl = CaCl2 + CO2 + H2O
The flask is made up to 250cm³. 20.0cm³ of this solution is then taken and titrated with 0.115 mol dm^-3 of sodium hydroxide.
What volume of sodium hydroxide is required for neutralisation???
A 0.4108g sample of calcium carbonate is added to a flask which contains 15.00cm³ of 2.00 mol dm^-3 hydrochloric acid.
CaCO3 +2HCl = CaCl2 + CO2 + H2O
The flask is made up to 250cm³. 20.0cm³ of this solution is then taken and titrated with 0.115 mol dm^-3 of sodium hydroxide.
What volume of sodium hydroxide is required for neutralisation???
Original post by kimmylee
Hey!
A 0.4108g sample of calcium carbonate is added to a flask which contains 15.00cm³ of 2.00 mol dm^-3 hydrochloric acid.
CaCO3 +2HCl = CaCl2 + CO2 + H2O
The flask is made up to 250cm³. 20.0cm³ of this solution is then taken and titrated with 0.115 mol dm^-3 of sodium hydroxide.
What volume of sodium hydroxide is required for neutralisation???
A 0.4108g sample of calcium carbonate is added to a flask which contains 15.00cm³ of 2.00 mol dm^-3 hydrochloric acid.
CaCO3 +2HCl = CaCl2 + CO2 + H2O
The flask is made up to 250cm³. 20.0cm³ of this solution is then taken and titrated with 0.115 mol dm^-3 of sodium hydroxide.
What volume of sodium hydroxide is required for neutralisation???
You need to explain what you don't understand, or what you have tried, first.
Original post by charco
You need to explain what you don't understand, or what you have tried, first.
n(CaCO3) = 0.4108/(40.1+12.(3x16)) = 0.00410mol
n(HCl)= 2.00x(15.00x10^-3) = 0.03mol
ratio 2:1 ???
n(NaOH) = 0.03/2= 0.015
V=n/c. = 0.015/0.115 = 0.130cm³. ??
Original post by kimmylee
Hey!
A 0.4108g sample of calcium carbonate is added to a flask which contains 15.00cm³ of 2.00 mol dm^-3 hydrochloric acid.
CaCO3 +2HCl = CaCl2 + CO2 + H2O
The flask is made up to 250cm³. 20.0cm³ of this solution is then taken and titrated with 0.115 mol dm^-3 of sodium hydroxide.
What volume of sodium hydroxide is required for neutralisation???
A 0.4108g sample of calcium carbonate is added to a flask which contains 15.00cm³ of 2.00 mol dm^-3 hydrochloric acid.
CaCO3 +2HCl = CaCl2 + CO2 + H2O
The flask is made up to 250cm³. 20.0cm³ of this solution is then taken and titrated with 0.115 mol dm^-3 of sodium hydroxide.
What volume of sodium hydroxide is required for neutralisation???
Second part equation: NaOH+HCl--> H20+NaCl
Find the concentration of HCl from: CaCO3 +2HCl = CaCl2 + CO2 + H2O.The number of moles you have now will be for 250cm3.
Therefore find the number of moles of HCl in 20cm3.
( 2nd part equation) Because its a 1:1 ratio the no. moles of HCl in the 20cm3 solution are the same as the no. moles of NaOH. Then use the equation to find the volume for NaOH and you should get the answer from there.
(edited 6 years ago)
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