Write complete ionic equation to show the reaction of aqueous lead(ii) nitrate with aqueous potassium sulfate to form solid lead(ii) sulfate and aqueous potassium nitrate.

Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺ (aq) + SO₄²⁻ (aq) → PbSO₄(s) + 2K⁺ + 2 NO₃⁻ (aq)

Explanation:

1) Start understanding the nature of the reactants, the kind of reaction that occurs, and the nature of the products:

i) word equation (given):

aqueous lead(II) nitrate + aqueous potassium sulfate → solid lead(II) sulfate + aqueous potassium nitrate

ii) molecular equation, including the phases:

Pb(NO₃)₂ (aq) + K₂SO₄ (aq) → PbSO₄(s) + 2KNO₃ (aq)

iii) It is a combination of two salts to form two new salts, in a double replacement reaction.

2) Separate the ionic compounds in solution into its ions

i) Reactant side:

Pb(NO₃)₂ (aq) = Pb²⁺ (aq) + 2NO₃⁻ (aq)

K₂SO₄ (aq) = 2K⁺ (aq) + SO₄²⁻ (aq)

ii) Product side

PbSO₄(s) remains unchanged since it did not ionize

2KNO₃(aq) = 2K⁺ + 2 NO₃⁻ (aq)

3) Write both sides, i.e. the whole reaction in the ionic form:

Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺ (aq) + SO₄²⁻ (aq) → PbSO₄(s) + 2K⁺ + 2 NO₃⁻ (aq)

That is the total ionic equation.

the answer is b. dilution

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer