How do i get the theoretical yeild and molar concentration of this.. startingout with 2 solutions you have?

Question

.800 M sodium phosphate, and the other is .800 lead(II) acetate if you pour 100 mL of the sodium phosphate and you take another 50.mL of the lead(II) acetate into the same beaker, how do you calculate the molar concentration and theoritical yeild?

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if anyone does help could you show me how cause I have a few more problems to do just like this one and I dont know how to calculate eithor! thanks

Answer 1

First write the balanced equation for the reaction of sodium phosphate and lead(II)acetate. The coefficient will tell you the molar ratios of the reactants and products.

liters x molarity = moles

0.100 liters of sodium phosphate x 0.800 M = 0.0800 mol

0.050 liters of lead(II)acetate x 0.800 M = 0.0400 mol

Now determine the limiting reactant.

Based on the moles of the limiting reactant, calculate the moles of lead(II)phosphate that should be formed.

Moles of lead(II)phosphate / 0.150 L = molarity

Moles of lead(II)phosphate x Molar Mass of lead(II)phosphate = grams of lead(II)phosphate.