CaCO3 (s) + H2SO4 (aq) --> CaSO4 (aq) + CO2 (g) + H2O(l)
This reaction can be replicated in the lab by performing a titration. If 7.21 g of CaCO3 solid is added to 1.20 L of 0.211 mol L- sulfuric acid, how may moles of acid will remain after the reaction?
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What volume of 0.411 mol L- KOH must be added to neutralize this excess acid?
2KOH (aq) + H2SO4 (aq) --> K2SO4 (aq) + 2H2O (l)
*Please tell me the steps and not just the answers because I have a lot of problems to do like this. Thanks! :) :D
This reaction can be replicated in the lab by performing a titration. If 7.21 g of CaCO3 solid is added to 1.20 L of 0.211 mol L- sulfuric acid, how may moles of acid will remain after the reaction?
--------------------------------------…
What volume of 0.411 mol L- KOH must be added to neutralize this excess acid?
2KOH (aq) + H2SO4 (aq) --> K2SO4 (aq) + 2H2O (l)
*Please tell me the steps and not just the answers because I have a lot of problems to do like this. Thanks! :) :D
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1) From the mass of CaCO3, find the mole of CaCO3 (n = m/MM). From the reaction, use the mole of CaCO3 you just found to find the mole of H2SO4. Hint: the reaction is 1 to 1 ratio, so the mole of H2SO4 is just the mole of CaCO3. From the Molarity and Volume of H2SO4 given, find the mole of H2SO4 (n = MV). Compare the mole obtained from the reaction with the one you just calculated and find the difference.
2) You just calculate the mole of H2SO4 remain in the solution, use it to find the mole of KOH that is needed to neutralize the acid. Hint: you need two mole of KOH for 1 mole of H2SO4. From the mole of KOH just calculated, find the volume of KOH (V = n/M)
2) You just calculate the mole of H2SO4 remain in the solution, use it to find the mole of KOH that is needed to neutralize the acid. Hint: you need two mole of KOH for 1 mole of H2SO4. From the mole of KOH just calculated, find the volume of KOH (V = n/M)