If 25 mL of a 0105 M HCI solution are added to 025 g of soli

If 25 mL of a 0.105 M HCI solution are added to 0.25 g of solid Caco, will any of the remain after the reaction is complete? so. caco, ? If so, how much? Show work to justify your answer A 0.950 g sample of iron net reaction can be writen as ore was dissolved and reacted with potassium permanganate. The 236 mL of a 00525M MnO, solution were required to react wit the Fe a. How many grams of iron were in the sample? b. What is the percent Fe in the sample?

Solution

Ans 5 :

Number of moles of HCl = 0.105 x 0.025 = 0.002625 mol

Number of moles of CaCO3 = 25 / molar mass

= 0.25 / 100.0869

= 0.00249 mol

Number of moles of CaCO3 utilised will be half of HCl = 0.002625 / 2 = 0.0013125 mol

So number of moles of CaCO3 left = 0.00249 - 0.0013125 = 0.0011775 mol

Mass of CaCO3 left = 0.0011775 x 100.0869

= 0.11785 g