Question

If 25 moles of a 6L solution of cesium hydroxide react with a 7M 2L solution of nitrous acid, how much water was created? What is the concentration of hydroxide ions after completion of the reaction?

Answer 1

we're looking for the moles of water in the products and the final concentration of (OH- ions) First we're going to set up a Acid-Base reaction which always forms water. CsOH + HNO2 -----> H2O + CsNO2 (Which is already balanced) Next we find the moles of HNO2 and identify our Limiting reactant (2L x 7mol/L = 14mol) this means our HNO2 is limiting. This also means we have a product of 14 moles of water (which is one of our answers). Now for every mole of water and CsOH we have 1 mole of hydroxide ions each. So we have a total of 25 moles of Hydroxide ions. We also have a total of 8 liters so we divide 25/8 = 3.125mol/L (OH-) (Which is our final concentration of Hydroxide ions.)