What volume of the solution in 5a would be necessary to react completely with 23.00mL of 0.549M NH3 according to the equation: NiSO4+6NH3 -> [Ni(NH3)6]SO4
The molarity of ta is 1.667 x 10^-4M
Is ta a typo for 5a? Is tghe solution is 5a(ta) NiSO4?
mols NH3 = M x L = ?
mols NiSO4 = (1/6) x mols NH3
M NiSO4 = mols NiSO4/L NiSO4. You know M and mols, solve for L.
To determine the volume of the solution in 5a required to react completely with 23.00 mL of 0.549 M NH3, we need to use the stoichiometry of the reaction and the given molarity of solution 5a.
First, let's examine the balanced equation:
NiSO4 + 6NH3 -> [Ni(NH3)6]SO4
From the balanced equation, we can see that the ratio between NiSO4 and NH3 is 1:6. This means that one molecule of NiSO4 requires 6 molecules of NH3 to react completely.
Next, we need to determine the number of moles of NH3 present in 23.00 mL of 0.549 M NH3.
We can use the formula:
moles = (volume in liters) x (molarity)
moles of NH3 = (23.00 mL / 1000 mL/L) x (0.549 mol/L)
= 0.012627 mol
Since the stoichiometry tells us that 6 moles of NH3 are required for every mole of NiSO4, we can now calculate the amount of NiSO4 required.
moles of NiSO4 = (0.012627 mol / 6)
= 0.002105 mol
Now, we can finally calculate the volume of solution 5a needed to react completely with the given amount of NiSO4.
volume of solution 5a = (moles of NiSO4) / (molarity of solution 5a)
volume of solution 5a = (0.002105 mol) / (1.667 x 10^(-4) mol/L)
= 12.625 L
Therefore, the volume of solution in 5a necessary to react completely with 23.00 mL of 0.549 M NH3 is 12.625 liters.