A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are 50.0 mL of 0.100 M hydrochloric acid, 100.0 mL of 0.200 M of nintric acid,500. mL of 0.0100 M calcium hydroxide, and 200.0 mL of 0.100 M rubidium hydroxide. Is the resulting solution neutral? If not, calculate the concentration of excess H+ or OH- ions left in solution.

mols HCl = M x L = ?

mols HNO3 = N x L = ?
Total mols H^+ \ sum

mols Ca(OH)2 = M x L = ?
mols OH^- = 2xmols Ca(OH)2
mols RbOH = M x L = ?
Total mols OH^- = ?
You can take it from there.

To determine whether the resulting solution is neutral or not, we need to calculate the concentration of excess H+ or OH- ions left in the solution.

Step 1: Let's start by writing down the balanced chemical equations for the reactions that occur between the acidic and basic solutions:

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) (1)
HNO3(aq) + Ca(OH)2(aq) → 2H2O(l) + Ca(NO3)2(aq) (2)
RbOH(aq) + HCl(aq) → RbCl(aq) + H2O(l) (3)
RbOH(aq) + HNO3(aq) → RbNO3(aq) + H2O(l) (4)

Step 2: Calculate the moles of each reactant:

Moles of HCl = Molarity × Volume (in L) = 0.100 M × 0.050 L = 0.005 mol
Moles of HNO3 = Molarity × Volume (in L) = 0.200 M × 0.100 L = 0.020 mol
Moles of Ca(OH)2 = Molarity × Volume (in L) = 0.0100 M × 0.500 L = 0.005 mol
Moles of RbOH = Molarity × Volume (in L) = 0.100 M × 0.200 L = 0.020 mol

Step 3: Determine the limiting reactant. This is the reactant that is completely consumed and determines the amount of product formed.

From the balanced chemical equations, we can see that both HCl and Ca(OH)2 react in a 1:1 ratio with NaOH and HNO3, respectively. However, RbOH reacts in a 1:2 ratio with HCl and HNO3. Since there are equal moles of HCl and RbOH (0.005 mol) and an equal number of moles of Ca(OH)2, the limiting reactant is HNO3 because there are twice as many moles of RbOH (0.020 mol) compared to HNO3.

Step 4: Calculate the moles of excess H+ or OH- ions:

From equation (4), we can see that for every 1 mole of HNO3, 2 moles of OH- are consumed. Therefore, with 0.020 mol of HNO3, only half of that (0.010 mol) will react to produce water. This means there is an excess of OH- ions.

Step 5: Calculate the concentration of OH- ions:

The total volume of the solution after mixing is 50.0 mL + 100.0 mL + 500.0 mL + 200.0 mL = 850.0 mL = 0.850 L

Concentration of OH- ions = Moles of OH- ions / Volume of solution
= (0.010 mol / 0.850 L)
≈ 0.0118 M

Therefore, the resulting solution is not neutral. It is basic with a concentration of approximately 0.0118 M OH- ions.