Answer:
38.8 g of PbCrO₄ are precipitated from the reaction
Explanation:
We have the following chemical reaction:
K₂CrO₄ (aq) + Pb(NO₃)₂ (aq) → 2 KNO₃ (aq) + PbCrO₄ (s)
where:
(aq) - aqueous
(s) - solid
Now we use the following formula to calculate the number of moles:
molar concentration = number of moles / volume (L)
number of moles = molar concentration × volume (L)
number of moles of K₂CrO₄ = 3 × 0.1 = 0.3 moles
number of moles of Pb(NO₃)₂ = 2 × 0.1 = 0.2 moles
We see from the chemical reaction that 1 mole of K₂CrO₄ will react with 1 mole of Pb(NO₃)₂ so 0.3 moles of K₂CrO₄ will react with 0.3 moles of Pb(NO₃)₂ but we have only 0.2 moles of Pb(NO₃)₂ available, so the limiting reactant is Pb(NO₃)₂. Taking this in account we devise the following reasoning:
if 1 moles of Pb(NO₃)₂ produces 1 mole of PbCrO₄
then 0.2 moles of Pb(NO₃)₂ produces X moles of PbCrO₄
X = 0.2 moles of PbCrO₄
number of moles = mass / molar weight
mass = number of moles × molar weight
mass of PbCrO₄ = 0.2 × 194 = 38.8 g
Learn more about:
molar concentration
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