Answered

Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Ask your questions and receive precise answers from experienced professionals across different disciplines. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

How many grams of lead (II) nitrate are needed to fully react 23.5 mL of 0.55 M sodium chloride in the precipitation of lead (II) chloride?

Sagot :

zcathy

Answer:

1.8 g of Pb(NO3)2

Explanation:

Find the moles of sodium chloride used by multiplying the molarity by the volume of sodium chloride.

Molarity = mol/L

Convert 23.5 mL to L.

23.5 mL x (1 L/1000 mL) = 0.0235 L

Multiply molarity by volume.

0.55 M = mol/0.0235 L

(0.55 M)(0.0235 L) = mol

mol = 0.012925

You have the moles of sodium chloride used so you can convert this to moels of lead (II) nitrate with stoichiometry. First, you need the balanced chemical equation.

Pb(NO3)2 + 2 NaCl -> PbCl2 + 2NaNO3

Convert 0.012925 mol NaCl with mole to mole ratio. In this case, it's 1:2.

0.012925 mol NaCl x (1 mol PbCl2/2 mol NaCl) = 0.0064625 mol PbCl2

Convert moles of PbCl2 to grams with molar mass.

0.0064625 mol PbCl2 x (278.10 g/1 mol) = 1.79722... g

Round to sig figs.

1.8 grams of PbCl2

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.