An aqueous solution containing 5.74 g of lead(ii) nitrate is added to an aqueous solution containing 6.26 g of potassium chloride to generate solid lead(ii) chloride and potassium nitrate. write the balanced chemical equation for this reaction. be sure to include all physical states.

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)

27. =3,500,000 x 4,000 = 14,000,000,000

The answer is just took the test