An aqueous solution containing 5.74 g of lead(ii) nitrate is added to an aqueous solution containing 6.26 g of potassium chlorid
e to generate solid lead(ii) chloride and potassium nitrate. write the balanced chemical equation for this reaction. be sure to include all physical states.
<span>Pb(NO3)2(aq) + 2 KCl(aq) ==> PbCl2(s) + 2 KNO3(aq)
Let's write the balanced equation
lead(II) nitrate is Pb(NO3)2
potassium chloride is KCl
lead(II) chloride is PbCl2
potassium nitrate is KNO3
So the basic equation is
Pb(NO3)2 + KCl ==> PbCl2 + KNO3
But it's not balanced. I see 2 nitrogen on the left and 1 nitrogen on the right. So add a coefficient to the KNO3, getting
Pb(NO3)2 + KCl ==> PbCl2 + 2 KNO3
Now there's only 1 potassium on the left, but 2 on the right. So add another coefficient, getting
Pb(NO3)2 + 2 KCl ==> PbCl2 + 2 KNO3
And it's balanced. Now to include the physical states.
Pb(NO3)2 is soluble in water, so it's state will be (aq)
KCl is also soluble in water, so it too will be (aq)
PbCl2 is poorly soluble in water. So it will become a solid precipitate. It's state will be (s)
KNO3 is quite soluble in water with it's solubility increasing rapidly with higher temperatures. So it's state will be (aq).
So let's add the states to the reactants and products.
Pb(NO3)2(aq) + 2 KCl(aq) ==> PbCl2(s) + 2 KNO3(aq)</span>
The reaction involves the heterolytic fission of the Br-Br bond in the bromine molecule to yield a bromine cation which attacks the but-1-ene to form a cyclic intermediate called the brominium ion. The bromine anion must now attack from the opposite face of the brominium ion due to steric clashes to form a product of a 1,2-dibromoalkane having the anti- stereochemistry.
