When a reaction mixture with a total volume of 1390 mL that is 0.00791 M aqueous Ba2+ was stoichiometrically produced as per the balanced equation, what mass (g) of Ba2+ was required?
Ba(OH)2(aq) + 2 HClO4(aq) → Ba(ClO4)2(aq) + 2 H2O(l)
.00791x1.39=mole of Ba2+
mole of Ba2+ xmolecular weight=grams Ba2+
To calculate the mass of Ba2+ required, we need to follow these steps:
1. Calculate the number of moles of Ba2+ in the reaction mixture.
The given concentration of Ba2+ is 0.00791 M. To find the number of moles, we can multiply the concentration by the total volume in liters:
0.00791 M * 1.39 L = 0.01099 moles Ba2+
2. Calculate the molar mass of Ba2+.
The molar mass of Ba2+ can be found by adding the atomic masses of each element in Ba2+. The atomic mass of Ba is 137.33 g/mol.
Molar mass of Ba2+ = (2 * atomic mass of Ba) = (2 * 137.33 g/mol) = 274.66 g/mol
3. Calculate the mass of Ba2+.
To calculate the mass of Ba2+ in grams, we can multiply the number of moles by the molar mass:
Mass of Ba2+ = 0.01099 moles * 274.66 g/mol = 3.02 grams
Therefore, the mass of Ba2+ required is 3.02 grams.