My textbook doesn't show me how to do problems like the following, and I'm having a hard time finding similar problems online. Most of the molarity problems I'm finding involve mixing a substance with water, but these involve mixing moles of substances with moles of other substances and I'm confounded as to how to try to solve them. If anyone can show me steps to solve these two, I would be extremely grateful:

1) If 35 mL of 2.0 M H2SO4 is mixed with 35mL of 1.0 M NaOH, how many moles of H2SO4 remain after completion of the reaction?

2) What is the molarity of Ca(NO3)2 in a solution resulting from mixing 150.0 mL of 0.0200 M HNO3 with 150.0 mL of 0.0100 M Ca(OH)2?

Thank you so much. :)

For this type of problem the usual approach is to start with a balanced equation for the reaction. This will tell you how many moles of each react together.

Then using the using the molarity and volume calculate the number of moles of each component.
For example the number of moles of H2SO4 in the first problem is
(35 x 2.0/1000)moles

Based on the equation and the number of moles of each compound work out which compound is in excess. In Q1 it is the H2SO4 as we are told this in the question, but you should convince yourself this is correct.

for 1) you can the find the number of moles (Mx) of H2SO4 left over (remain). The question does not ask you to calculate the final concentration!

for 2) the compund that is not in excess will determine the number of moles of Ca(NO3)2 in solution (Mc). The new volume is 300.0 ml so the molarity is

Mc/0.3000

Hope this helps.

To solve these types of problems, you can use the concept of stoichiometry and the definition of molarity. Let me explain how to solve each problem step by step.

1) If 35 mL of 2.0 M H2SO4 is mixed with 35 mL of 1.0 M NaOH, you can start by writing the balanced chemical equation for the reaction:

H2SO4 + 2 NaOH -> Na2SO4 + 2 H2O

From the equation, you can see that 1 mole of H2SO4 reacts with 2 moles of NaOH to produce 1 mole of Na2SO4 and 2 moles of water. In this case, the reaction will go to completion because both reactants are strong electrolytes.

To solve the problem, you need to determine the limiting reactant. Since the volume of both solutions is the same, you can assume that the molarities of H2SO4 and NaOH are the same after mixing. Therefore, to calculate the limiting reactant, you can compare the number of moles of H2SO4 with NaOH.

First, calculate the number of moles of H2SO4:
Moles of H2SO4 = molarity x volume = 2.0 M x 0.035 L = 0.07 moles

Next, calculate the number of moles of NaOH:
Moles of NaOH = molarity x volume = 1.0 M x 0.035 L = 0.035 moles

Since you have twice as many moles of H2SO4 compared to NaOH, NaOH is the limiting reactant. This means that all of the NaOH will react, and H2SO4 will be left over.

To determine the remaining moles of H2SO4, you can subtract the moles of NaOH consumed from the initial moles of H2SO4:
Remaining moles of H2SO4 = initial moles - moles of NaOH consumed
= 0.07 moles - 0.035 moles
= 0.035 moles

Therefore, after the completion of the reaction, 0.035 moles of H2SO4 will remain.

2) To find the molarity of Ca(NO3)2 in the solution resulting from mixing 150.0 mL of 0.0200 M HNO3 with 150.0 mL of 0.0100 M Ca(OH)2, you will again use the concept of stoichiometry.

First, calculate the number of moles of HNO3:
Moles of HNO3 = molarity x volume = 0.0200 M x 0.150 L = 0.003 moles

Next, calculate the number of moles of Ca(OH)2:
Moles of Ca(OH)2 = molarity x volume = 0.0100 M x 0.150 L = 0.0015 moles

The balanced chemical equation for the reaction between HNO3 and Ca(OH)2 is:

2 HNO3 + Ca(OH)2 -> 2 H2O + Ca(NO3)2

From the equation, you can see that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of water and 1 mole of Ca(NO3)2.

To determine the moles of Ca(NO3)2 formed, you can compare the moles of HNO3 with Ca(OH)2. Since the reaction uses a 1:1 stoichiometric ratio, the moles of Ca(NO3)2 formed will be the same as the moles of Ca(OH)2 used. Therefore, the moles of Ca(NO3)2 formed are 0.0015 moles.

Now, calculate the final volume of the solution by adding the volumes of the two solutions:
Final volume = 150 mL + 150 mL = 300 mL = 0.300 L

Finally, calculate the molarity of Ca(NO3)2 in the solution:
Molarity of Ca(NO3)2 = Moles of Ca(NO3)2 / Final volume
= 0.0015 moles / 0.300 L
= 0.005 M

Therefore, the molarity of Ca(NO3)2 in the resulting solution is 0.005 M.

I hope this helps! Let me know if you have any other questions.