Mass of reactant - Stuck

Question

I am having trouble calculating moles and using the given information to answer the following questions.

2AgNO₃(aq) K₂CrO₄(aq) --> Ag₂CrO₄(s) 2KNO₃(aq)

0.778g of precipitate is formed.

1. What is the mass of potassium chromate that reacted?

2. What is the mass of silver nitrate that reacted?

Answering either one would be sufficient I guess as I would understand it.

Thanks so very much for those who take the time to explain.

Answer 1

In this situation, ur ppt is Ag₂CrO. This is denoted by the s in the eqn.

So firstly u gotta find the molecular mass of Ag₂CrO₄.

this is calculated by adding up the mass of diff elements given in the periodic table. diff  tables differ by a lil.

thus Ag₂CrO₄ is ( 108x2)+ (52)+(16x4)=332g

this denotes 1 mole of Ag₂CrO₄. so now we have to calculate how many moles of Ag₂CrO₄ is derived in this reaction

0.778/332=0.00234 moles

Next u find the molecular mass of K₂CrO₄which is 194.2g.

To find the mass of K₂CrO₄used, we look at the mole ratio of the eqn, which is 1:1

Thus the mass of K₂CrO₄ tt reacted will be 194.2x0.00234= 0.455g

This method is used for the next q too. just remember to use the mole ratio of 2:1.

This method is used for every other calculations.

Good luck calculating n i hope this helps...=)

Answer 2

Mm(Ag2CrO4) = 332 g mol-1

So 0.778/332 = 0.0023 mol of silver chromate were precipitated.

So (from balanced equation) there were 0.0023 mol of potassium chromate(VI used and 0.0046 mol of silver nitrate.

Masses can be calculated from these values by multiplying by relevant molar mass.