A0.450 g0.450 g sample of impure caco3(s)caco3(s) is dissolved in 50.0 ml50.0 ml of 0.150 m hcl(aq)0.150 m hcl(aq) . the equation for the reaction is caco3(s)+2hcl(aq)⟶cacl2(aq)+h2o(l)+ co2(g) caco3(s)+2hcl(aq)⟶cacl2(aq)+h2o(l)+ co2(g) the excess hcl(aq)hcl(aq) is titrated by 9.35 ml9.35 ml of 0.125 m naoh(aq)0.125 m naoh(aq) . calculate the mass percentage of caco3(s)caco3(s) in the sample.

The mass percentage of CaCO3 in the sample is 70.4%

Explanation:

Step 1: Data given

mass of the sample = 0.450 g

volume of the 0.150 M HCl = 50.0 mL = 0.05L

The excess HCl is titrated by 9.35 mLof 0.125 M NaOH

Molar mass CaCO3 = 100.0869 g/mol

Step 2: The balanced equation

CaCO3(s)+2HCl(aq) → CaCl2(aq)+H2O(l)+CO2(g)

Step 3: Calculate moles of NaOH

Number of moles NaOH = molarity NaOH * Volume

Number of moles NaOH = 0.125 M * 0.00935 L

Number of moles of NaOH = 0.00116875 moles

Step 4: Calculate moles of HCl

For 1 mole of NaOH consumed, we need 1 mole of HCl to produce 1 mole of NaCl and 1 mole of H2O

For 0.00116875 moles of NaOH, we have 0.00116875 moles moles of HCl

in excess

Step 5: Calculate moles of HCl reacted with CaCO3

Number of moles = molarity HCl * Volume

Number of moles = 0.150 M *0.05L = 0.0075 moles

Moles of HCl reacted with CaCO3 = 0.0075 - 0.00116875 = 0.00633125 moles

Step 6: Calculate moles of CaCO3

For 2 moles of HCl consumed, we need 1 mole of CaCO3 to produce 1 mole of CaCl2 and 1 mole of H2O

For 0.00633125 moles of HCl, we have 0.00633125/2 = 0.003165625 moles

Step 7: Calculate mass of CaCO3

mass CaCO3 = number of moles of CaCO3 / molar mass of CaCO3

mass of CaCO3 = 0.003165625 moles * 100.0869 g/mol = 0.317 grams

Step 8: Calculate the mass percentage of CaCO3

(0.317 grams / 0.450 grams) * 100 % = 70.4 %

The mass percentage of CaCO3 in the sample is 70.4%

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.