A 0.450-gram sample of impure CaCO_3(s) is dissolved in 50.0 mL of 0.150 M HCI(a

Question

A 0.450-gram sample of impure CaCO_3(s) is dissolved in 50.0 mL of 0.150 M HCI(aq). The equation for the reaction is The excess HCI(aq) is titrated by 4.80 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCC_3(s) in the sample. You can determine the number of moles of excess HCI from the amount of NaOH needed to titrate it. Subtract that value from total number of moles of HCI (50.0 mL of 0.150 M HCI). The result is the number of moles of HCI that reacted with CaCO_3. Once you know the number of moles of HCI that reacted, use the stoichiometry of the chemical equation to find the amount of CaCO_3 that reacted. Finally, express that amount of CaCO_3 as a mass percentage of the original sample.

Explanation / Answer

Moles of HCl initailly present = 0.15*50/1000=0.0075 moles

Stoichiometric ratio of reactants = 1:2

Actual molar ratio of reactants = 0.0045:0.0075= 1:1.67

The reaction between HCl and NaOH is given by

HCl+NaOH-à NaCl+H2O

Moles of NaOH used to titrate excess HCl= 0.125*4.8/1000=0.0006

Moles of HCl is same as that of moles of NaOH= 0.0006

Moles used to react with CaCO3 in the sample =moles initially- excess moles of HCl= 0.0075-0.0006=0.0069 moles

CaCO3 Consumesd =0.5 times moles of HCl =0.5*0.0069=0.00345 moles

Mass of CaCO3= moles of CaCO3* molecular weight= 0.00345*100=0.345gm

Percentage of CaCO3 in the sample = 100*0.345/0.45=76.67%

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