Question

How much lead(II) chromate (PbCrO4) is precipitated by mixing 100 mL of aqueous 3 M

Answer 1

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

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molar concentration

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