I am getting a stupid answer for this question. Can someone tell me where I've gone wrong or talk me through the right answer? Thanks
Question: A sample of chalk weighing 0.135g was reacted with 40.00cm3 of 1mol/l hydrochloric acid (an excess). After the reaction was complete the solution was transferred to a 250cm3 volumetric flask and made up to the mark with distilled water. 25.00cm3 portions of this solution were then titrated against 0.15mol/l sodium hydroxide using methyl orange as the indicator. An average of 25cm3 of the sodium hydroxide was required. Calculate the percentage purity of the chalk.
My Answer: HCl + NaOH --> H2O + NaCl 1 mol HCl reacts with 1 mol NaOH No. moles NaOH = 0.15 x 0.025 = 0.00375mol 0.00375 mol HCl reacts with 0.00375 mol NaOH
0.00375 mol in 1/10 of reacted solution 0.0375 mol in reacted solution No. moles HCl = 0.04 x 1 = 0.04 mol No. moles HCl reacted = 0.04 - 0.037 = 0.003 mol
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O 1 mol CaCO3 reacts with 2 mol HCl 0.0015 mol reacts with 0.003 mol HCl
1 mol CaCO3 = 100g 0.0015 mol = 0.15g
This would give 0.15/0.135x100 which is 111.11% purity. This is stupid
Hi, When you did the subtraction for the moles of HCL, you left out a decimal place. put it back in and there are: 0.0025moles of HCL reacted /2 =0.00125moles of CaCO3 on the 0.135g sample =0.125g of calcium carbonate in the sample % purity = 0.125/0.135 x 100 =92.6% pure (3sf) Hope that helps!
I am getting a stupid answer for this question. Can someone tell me where I've gone wrong or talk me through the right answer? Thanks
Question: A sample of chalk weighing 0.135g was reacted with 40.00cm3 of 1mol/l hydrochloric acid (an excess). After the reaction was complete the solution was transferred to a 250cm3 volumetric flask and made up to the mark with distilled water. 25.00cm3 portions of this solution were then titrated against 0.15mol/l sodium hydroxide using methyl orange as the indicator. An average of 25cm3 of the sodium hydroxide was required. Calculate the percentage purity of the chalk.
My answer:
Total moles of HCl = 0.04 x 1 = 0.04 mol Moles of NaOH in the titre = 0.025 x 0.15 = 0.00375 Therefore moles of NaOH requiired to neutralise all of the excess acid = 0.0375 mol Acid used up in the reaction = 0.04 - 0.0375 = 0.0025 mol
Total moles of HCl = 0.04 x 1 = 0.04 mol Moles of NaOH in the titre = 0.025 x 0.15 = 0.00375 Therefore moles of NaOH requiired to neutralise all of the excess acid = 0.0375 mol Acid used up in the reaction = 0.04 - 0.0375 = 0.0025 mol