Question

1)How many moles of HCl are in 67 mL of 0.11 M HCl?

2)What mass of calcium carbonate is needed for complete reaction with the HCI in the previous question

HCl in (a)?Calcium carbonate reacts with HCl according to the following equation:
2HCl(aq)+CaCO3(s)→CaCl2(aq)+H2O(l)+CO2(g)

3.Aqueous ammonia is commercially available at a concentration of 16.0 M.

How much of the concentrated solution would you use to prepare 600.0 mL of a 1.10 M solution?

4a. How many grams are in 1

Eq of the Mg2+?

4b. How many grams in 1 mEq Mg2+?

How many grams are in 1 Eq of the SO2−4?

How many grams in 1 mEq SO42−?

How many grams are in 1 Eq of the Al3+?

How many grams in 1 mEq Al3+?

How many grams in 1 Eq PO3−4?

How many grams in 1 mEq PO3−4?

Answer 1

1.

use the formula

Moles = Molarity of the solution x volume of the solution in liters

given volume is 67 mL = 0.067 L

moles = 0.11M x 0.067 L = 0.00737 mol

2.

lets write the balanced equation

CaCO3 (s) + 2HCl (aq) ---> CaCl2 + H2O (l) + CO2 (g)

from the balanced equation

1 mol of HCl required half mole of CaCO3; accordingly

0.00737 mol of HCl required 0.00737 mol / 2 = 0.003685 mol CaCO3

mass of CaCO3 = moles x molar mass = 0.003685 mol x 100.0869 g/mol = 0.37 g