In this reaction: Mg (s) + I₂ (s) → MgI₂ (s), if 10.0 g of Mg reacts with 60.0 g of I₂, and 53.88 g of MgI₂ form, what is the pe
2 answers:
We know the law of conservation of mass
- It states that mass is neither formed nor destroyed in any chemical reaction.
- Mass of reactants=Mass of products.
Here
- Mg and I_2 are reactants
- MgI_2 is product with some yield.
- Mass of reactants=10+60.0=70.0g
- Mass of MgI_2=53.88g
- Mass of yield=Product-MgI_2=70-53.88=16.12g
Lets find the percentage
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Answer:
this is the concentration required to initiate precipitation
Explanation:
⇄
Precipitation starts when ionic product is greater than solubility product.
Ip>Ksp
Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.
This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.
lets solubility=S
this is the concentration required to initiate precipitation