Question

A student is running an experiment in which 73.4 grams of BaI2 is needed, but the only jar of reagent in the lab is labelled barium iodide dihydrate. How many grams of the hydrate must the student weigh out in order to get the desired amount of the anhydrous compound?

1. How many GRAMS of potassium are present in 1.73 grams of potassium chromate, K₂CrO₄ ?
grams potassium.

How many GRAMS of potassium chromate can be made from 2.35 grams of potassium ?
grams potassium chromate.

Answer 1

Barium iodide dihydrate is BaI2.2H2O

Molar mass of BaI2.2H2O=Molar mass of Ba+2xmolar mass of I+4xmolar mass of H+2xmolar mass of O

=137.3 g/mol+2x126.9 g/mol+4x1 g/mol+2x16 g/mol

=137.3 g/mol+253.8 g/mol+4 g/mol+32 g/mol

=427.1 g/mol

Molar mass of BaI2=Molar mass of Ba+2xmolar mass of I

=137.3 g/mol+2x126.9 g/mol

=137.3 g/mol+253.8 g/mol

=391.1 g/mol

1 mol BaI2 is present in 1 mol BaI2.2H2O

This means 391.1 g BaI2 is present in 427.1 g BaI2.2H2O

So 1 g BaI2 is present in 427.1 g/391.1 g BaI2.2H2O

73.4 g BaI2 is present in (427.1 g/391.1 g)x73.4 g= 80.2 g BaI2.2H2O

So the student needs to weigh out 80.2 g BaI2.2H2O (the hydrate) to obtain 73.4 g BaI2.

1. Potassium chromate is K2CrO4.

Molar mass of K2CrO4=2xMolar mass of K+Molar mass of Cr+4xmolar mass of O

=2x39 g/mol+52 g/mol+4x16 g/mol

=78 g/mol+52 g/mol+64 g/mol

=194 g/mol

Molar mass of K=39 g/mol

Number of moles in 1.73 g K2CrO4=Given mass/molar mass

=1.73 g/194 g/mol=0.00892 mol

1 mol K2CrO4 contains 2 mol K.

So 0.00892 mol K2CrO4 contains 2 x 0.00892 mol = 0.01784 mol K

Mass of 0.01784 mol K=number of moles x molar mass of K

=0.01784 mol x 39 g≈0.70 g

So 1.73 g K2CrO4 contains 0.70 g K

Likewise

2 mol K can be used to make 1 mol K2CrO4

i.e. 78 g K can be used to make 194 g K2CrO4.

1 g K can be used to make (194 g/78 g) K2CrO4

So 2.35 g K can be used to make (194/78)x2.35 g=5.84 g K2CrO4