- #1
thepoll
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Hi, I've been trying and trying but I can't seem to solve this. Hopefully, there's someone here kind enough to help me through :X
2.5g of impure CaCO3 was dissolved in 25.0cm^3 of 2.00 mol dm^-3 HCL in a 100cm^3 volumetric flask and made up to the mark with distilled water. This solution is labelled as Solution X. 15.0cm^3 of this mixture required 20.0cm^3 of 0.0500 mol dm^-3 NaOH for neutralization.
Calculate the number of moles of HCL that has reacted with 2.5g of CaCO3.
1. I calculated the mole of HCL in 25.00cm^3 of 2.00 mol dm^-3 solution and also the number of moles of NaOH required to be titrated with 25cm^3 of the mixture of X.
2. I then tried calculating the concentration of Solution X that contains excess HCL. However, I got stuck there. If anyone could just hint me the next step, I would greatly appreciate it! thanks:)
2.5g of impure CaCO3 was dissolved in 25.0cm^3 of 2.00 mol dm^-3 HCL in a 100cm^3 volumetric flask and made up to the mark with distilled water. This solution is labelled as Solution X. 15.0cm^3 of this mixture required 20.0cm^3 of 0.0500 mol dm^-3 NaOH for neutralization.
Calculate the number of moles of HCL that has reacted with 2.5g of CaCO3.
1. I calculated the mole of HCL in 25.00cm^3 of 2.00 mol dm^-3 solution and also the number of moles of NaOH required to be titrated with 25cm^3 of the mixture of X.
2. I then tried calculating the concentration of Solution X that contains excess HCL. However, I got stuck there. If anyone could just hint me the next step, I would greatly appreciate it! thanks:)
