What mass of 95% pure CaCO3 will be require neutralise 50 mL of 0.5M HCl solution according to the following reaction?
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l) [Calculate upto second place of decimal point]
What mass of 95% pure CaCO3 will be require neutralise 50 mL of 0.5M HCl solution according to the following reaction?
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l) [Calculate upto second place of decimal point]
Moles of HCl = (50 / 1000) * 0.5 = 0.025 moles
So, moles of CaCO? used = 0.025 / 2 = 0.0125 moles = 1.25 g
95% (Total mass of CaCO? ) = 1.25 g
Total mass of CaCO? = 1.25 / 0.95 = 1.32 g
Similar Questions for you
0.01 M NaOH,
M = 1 * 10-2

pOH = 2
pH = 2
Kp = Kc (RT)Dng
36 * 10–2 = Kc (0.0821 * 300)–1
Kc = 0.36 * 0.0821 * 300 = 8.86 » 9
A(g) ->B(g) + (g)
Initial moles n 0 &nbs
On increasing pressure, equilibrium moves in that direction where number of gaseous moles decreases.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
Learn more about...

Chemistry Ncert Solutions Class 11th 2023
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
See what others like you are asking & answering

