Question

1. Calculate the molarity of the following solutions.

87.0 g of HCl in 3.00 L of a HCl solution.

27.0 g of NaOH in 300. mL of a NaOH solution

2.Write the net ionic equation to show the formation of a precipitate (insoluble salt) when the following solutions are mixed. Write noreaction if there is no precipitate.

AgNO3(aq) and LiF(aq)

Na3PO4(aq) and Pb(NO3)2(aq)

Li2SO4(aq) and BaCl2(aq)

Express your answer as a chemical equation. Identify all of the phases in your answer. Enter noreaction if no precipitate is formed.

Answer 1

1. Calculate the molarity of the following solutions.

87.0 g of HCl in 3.00 L of a HCl solution.

H = 1.00794g/mol

Cl = 35.4527g/mol

M.W = 1.00794 + 35.4527 = 36.46064g/mol

No of moles==87./ 36.46064 = 2.386 mol

molarity: = molarity = moles of solute / liters of solution
=2.386 /3 = 0.7953M ------------------answer

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27.0 g of NaOH in 300. mL of a NaOH solution

Molar mass =NaOH= 39.997 g/mol=40g/mol

No of moles==27 /40=0.675 mol

molarity: = molarity = moles of solute / liters of solution

= (27 / 40) / 0.300 = 2.25M------------------answer

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AgNO₃(aq) + LiF(aq) = AgF(s) + LiNO₃(aq)
Reaction type: double replacement

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3 Pb(NO₃)₂(aq) + 2 Na₃PO₄(aq) = Pb₃(PO₄)₂(s) + 6 NaNO₃(aq)
Reaction type: double replacement

(Na3PO4 + Pb(NO3)2):

Balanced equation:

2Na3PO4(aq) + 3Pb(NO3)2 (aq) à 6NaNO3(aq) + Pb3(PO4)2(s)

Total Ionic equation:

6Na+ (aq) + 2PO4 3- (aq) + 3Pb 2+ (aq) + 6NO3- (aq) à 6Na +(aq) + 6NO3-(aq) +Pb3(PO4)2

Net ionic equation:

2PO4 3- (aq) + 3Pb 2+ (aq) à Pb3(PO4)2

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Li₂SO₄(aq) + BaCl₂(aq) = BaSO₄(s) + 2 LiCl(aq)
Reaction type: double replacement