Cement is primarily composed of calcium silicate compounds whose formulas can be

Question

Cement is primarily composed of calcium silicate compounds whose formulas can be represented as (CaO)x·SiO2. When cement is mixed with water and sand, it forms concrete. An important factor in the physical characteristics of concrete is its cement content. During highway construction, an important quality control test is to routinely sample the concrete mix and determine its CaO content. A 75.15 g concrete sample was treated with a 150.00 mL portion of 2.021 M standard HCl. The HCl reacts with the CaO of the cement and dissolves it according to the equation: CaO + 2 HCl CaCl2 + H2O After the reaction is complete, the sample was stirred thoroughly. From this sample, a 25.00 mL aliquot was removed and placed into an Erlenmeyer flask. Phenolphalein is added and the excess HCl was titrated with 1.080 M NaOH. A volume of 42.04 mL of NaOH was required to reach the endpoint. Calculate the mass percent of CaO in the concrete sample in the following steps. Hint: Remember to take the aliquot into account in your calculations. In the part (a) answer, you are calculating the amount of excess HCl titrated in the 25 mL aliquot. In the part (b) answer, you are calculating the quantity of excess HCl in the original 150 mL reaction solution volume. These quantities are related by a factor of 150/25.

(a) Calculate the concentration of excess HCl titrated by NaOH in the reaction mixture aliquot. _____ M HCl

(b) Calculate the number of moles of excess HCl that remain in the original solution after reaction with CaO.______ mol HCl (excess left after reaction w/CaO)

(c) Calculate the number of moles of HCl that reacted with CaO._______mol HCl (reacted w/CaO)

(d) Calculate the mass of CaO in the sample. Incorrect: Your answer is incorrect. g CaO (e) Calculate the mass percent CaO in the concrete.______ % CaO

Explanation / Answer

Ans. #a: Calculate excess [HCl] from Titration

At titration end point-          M1V1 (NaOH) = M2V2 (HCl)

            So, [HCl], M2 = (1.080 M x 42.04 mL) / 25.00 mL = 1.816 M

Hence, excess [HCl] = 1.816 M

#b. Moles of excess HCl = Excess [HCl] x Vol. in liter

                                                = 1.816 M x 0.150 L = 0.2724 mol

#c. Total Initial moles of HCl taken = Initial [HCl] x Vol. in liters

                                                = 2.021 M x 0.150 L = 0.30315 mol

Now,

            Moles of HCl consumed = Initial moles – Excess remaining moles

                                                = 0.30315 mol – 0.2724 mol = 0.03075 mol

Therefore, moles of HCl reacted with CaO = 0.03075 mol = Moles of HCl consumed

#d. Following stoichiometry of balanced reaction, 2 mol HCl reacts with 1 mol CaO.

So,

            Moles of CaO in sample = ( ½ ) x Moles of HCl

                                                            = ( ½ ) x 0.03075 mol = 0.015375 mol

# Mass of CaO in sample = Moles x MW = 0.015375 mol x 56.0774 g mol-1 = 0.863 g

#e. % CaO in sample = (Mass of CaO / Mass of sample) x 100

                                    = (0.863 g / 75.15 g) x 100

                                    = 1.15 %

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