How many grams of potassium nitrate (kno3) would form if 2.25 liters of a 1.50 molar lead nitrate pb(no3)2 solution reacts with 1.15 liters of a 2.75 molar potassium chromate k2cro4 solution? (2 points) pb(no3)2 + k2cro4 yields pbcro4 + 2kno3

First we must understand the balanced chemical equation:
Pb(NO3)2 + K2CrO4 ==> PbCr04 + 2KNO3

This shows us that two moles of potassium nitrate are formed from 1 mole of lead nitrate or potassium chromate solution. The next step is to find out how many moles of each reactant there are. Note the word Molar is a concentration that simply means moles per liter.

2.25L of 1.5M lead nitrate = 2.25x1.5 = 3.375 moles of lead nitrate
1.15L of 2.75M potassium chromate = 1.15x2.75 = 3.1625 moles

The important part here is to see that the number of moles of the reactants are different. We know the number of moles of products will be dependent on the number of moles of reactants, and in this case there is less potassium chromate than there is lead nitrate, so this is the limiting factor as there is a one to one relationship with both reactants. Therefore, the number of moles of potassium nitrate produced is 2 x number of moles of potassium chromate. i.e. 6.325 moles of potassium nitrate is liberated.

To work out the number of grams, we must find the molar mass (the mass of one mole) of KNO3, which is the sum of the molar mass of each of its component atoms that make up the molecule. I've looked this up as 101.1 grams per mole.

Now we simply times the molar mass by the number of moles to yield the final grams liberated: 6.325 moles  x 101.1 grams/mole = 639.4 grams of potassium nitrate is liberated from this reaction.

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