Question

When aqueous solutions of (NH4)2CrO4 and Ba(NO3 )2 are combined, BaCrO4 precipitates. Calculate the mass, in grams, of the BaCrO4 produced when 1.38 mL of 0.123 M Ba(NO3 )2 and 3.7 mL of 0.678 M (NH4)2CrO4 are mixed. Calculate the mass to 3 significant figures.

Answer 1

The chemical reaction of (NH4)2CrO4 and Ba(NO3 )2 as follows:

(NH4)2CrO4 + Ba(NO3)2 > BaCrO4 + 2 NH4NO3

Now calculate the moles of both reactants:

1.38 mL of 0.123 M Ba(NO3 )2 = 1.6974*10^-4 mol Ba(NO3 )2

3.7 mL of 0.678 M (NH4)2CrO4= 2.5086 *10^-3 mol (NH4)2CrO4

Here an excess of (NH4)2CrO4,

The limiting agent is Ba(NO3)2.

Now calculate the moles of BaCrO4 which is produced from 1.6974*10^-4 mol Ba(NO3 )2

1.6974*10^-4 mol Ba(NO3 )2 *1.0 mol BaCrO4 /1.0 mol Ba(NO3 )2

= 1.6974*10^-4 mol BaCrO4

Now convert the mole into grams with its molar mass:

1.6974*10^-4 mol BaCrO4* 253.37 g/mol

= 0.0430 g BaCrO4