You weigh out pure Calcium carbonate.... 0.1740 g a) Calculate how many whole dr

Question

You weigh out pure Calcium carbonate.... 0.1740 g a) Calculate how many whole drops (.05 mL/drop) of 6 M HCl are required to dissolve this. (I found this to be 11.59 drops or 12 full drops) How do you do part b? CaCO3 + 2HCl---> Ca2+ +2Cl- +H2O+CO2 please use 4 sig fig in all calculations b) What is the Ca concentration when this is dissolved and diluted to 100.00mL?

Explanation / Answer

moles of CaCO3 = 0.174/100 CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l) Each mole requires two moles HCl so moles of HCl = 2*0.174/100 a) V * 6 = 2*0.174/100 V = 5.8 x10^-4 L = 0.58 mL drops = 0.58/0.05 = 11.6 so 12 drops required b)moles of Ca = moles of CaCO3 = 0.174/100 conc = moles / volume = 0.174/(100*0.1) = 1.74 x 10^-3 M

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.