Question

What mass of CaCO3 is required to react completely with 0.75 M HCl???
PLS ANSWER THE QUESTION.........

Answer 1

N * V = # of moles

and then multiply the number of moles by the molecular weight of CaCO3 (100.09 g/mol).

You also need to consider the neutralization reaction that is occurring:

2 HCl + CaCO3 → CaCl2 + CO2 + H2O

So, for each CaCO3, you will consume to equivalents of HCl.

OK, time to solve the problem. The number of moles of HCl you have in 25 mL of 0.75 N HCl solution is:

0.025 L * 0.75 moles/L = 0.01875 moles.

Since you need twice as much HCl as CaCO3, divide this number (O.01875) by 2 and you get that it will neutralize 0.009375 moles of CaCO3. Now multiply by the molecular weight of CaCO3 to get:

0.009375 moles * 100.09 g/mole CaCO3 = 0.9383 g of CaCO3