Question

# If 25 moles of a 6L solution of cesium hydroxide react with a 7M 2L solution...

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?

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.)

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