Question

What volume of 12.0 M nitric acid is required to prepare 4.63 L of 0.200 M nitric acid ?

Determine the concentration (molarity) of a solution made by dissolving 35.5 g of sodium sulfide in 550.0 mL of solution

Consider the following reaction : 25.0mL of a 0.130 M sodium sulfate solution is reacted with 35.0 mL of a 0.124 M lead (ll) nitrate solution. The solvent for both solutions is water. If the reaction is carried out, what mass of lead (ll) sualfate eould be produced?

Write balanced equation

Give the states of all reactants and products and consider solubility rules

Determine the limiting reactant

For the nitrogen fixation reaction 3H2 + N2 <=> 2NH3

K=6.0 x 10^-2 at 500 c. If 0.250 M H2 and 0.050 M NH3 are present at equilibrium, what is the equilibrium concentration of N2 ?

Answer 1

1) Let for the HNO3 solution to be prepared, M1 = 0.200 M, V1 = 4.63 L

And for Stock HNO3 solution, M2 = 12.0 M and V2 = ?

By law of dilution,

M2*V2 = M1*V1

12.0 M * V2 = 0.200 M * 4.63 L

V2 = 0.200 M * 4.63 L / 12.0 M

V2 = 0.0771 L

V2 = 77.1 mL

77.1 mL of HNo3 needed.

77.1 mL of 12.0 M HNO3 when diluted to 4.63 L its concentratio will be 0.200 M

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2) Given mass of sodium sulphide Na2S = 35.5 g

Molar mass of Na2S = 78.05 g/mol

So, # of moles of Na₂S in 35.5 g = Given mass / Molar mass = 35.5 g / 78.05 gmol^-1. = 0.455 mol.

Voluume of the solution formed = 550.0 mL = 0.550 L

Molarity of Na2S i.e. [Na2S] = # of mles of Na2S / Volume of the solution in L = 0.455 mol / 0.550 L = 0.827 M

Molarity of the solution thus formed = 0.827 M

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