Answer:
Lead chromate form = 0.97 g
Explanation:
The Balanced chemical equation of the given reaction is as follows
K₂CrO₄ + Pb(NO₃)₂ → PbCrO₄ + 2 KNO₃
Given that
Mass of Pb(NO₃)₂ = 1 g
[tex]Mole=\frac{Mass}{Molar mass} = \frac{1}{331.2}= 0.003[/tex]
Mole of 25 ml of 1.00 (M) K₂CrO₄
Mole = 25 x 1 milimol = 25 x 10⁻³
Using mole ratio method to find limiting reagent
[tex]\frac{Mole}{Stoichiometry}[/tex]
[tex]For K_{2}CrO_{4} = \frac{25 X10^{-3} }{1}=25 X10^{-3} \\\\For Pb(NO_{3} )_{2} = \frac{0.003}{1}=0.003[/tex]
Hence K₂CrO₄(potassium chromate) is a limiting reagent.
So lead chromate formed = mole X molar mass
= 0.003 X 323.19 g
= 0.97 g
∴ 0.97 g lead chromate form when a 1.00-g sample of Pb(NO₃)₂ is added to 25.0 ml of 1.00 M K₂CrO₄ solution.
0.96 g of lead chromate would be formed.
From the balanced equation of the reaction:
K2CrO4 + Pb(NO3)2 → PbCrO4 + 2 KNO3
Mole ratio of Pb(NO3)2 to K2CrO4 = 1:1
Mole of 1.00 g Pb(NO3)2 = 1/331.2
= 0.003 mole
Mole of 25 mL, 1 M K2CrO4 = 0.025 x 1
= 0.205 mole
Limiting reactant = Pb(NO3)2
Mole ratio of Pb(NO3)2 and PbCrO4 = 1:1
Hence, equivalent mole of PbCrO4 = 0.003 mole
Mass of 0.003 mole PbCrO4 = 0.003 x 323.19
= 0.96 g
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