Help me answer this!!!!

Question

You have reacted 100.00 mL of 0.949M calcium nitrate solution with 100.00 mL of 0.955M sodium phosphate solution. After filtering and drying the precipitate, you find that you have 8.630g of calcium phosphate solid. What is your percent yield? Report your answer to 1 decimal place.

Answer 1

Write the correctly balanced equation:

3Ca(NO₃)₂(aq) + 2Na₃PO₄(aq) ==> Ca₃(PO₄)₂(s) + 6NaNO₃(aq) ... balanced equation

Next, find the limiting reactant. Easy way is divided moles of each reactant by its coefficient.

For Ca(NO₃)₂: 100.00 ml x 1 L / 1000 ml x 0.949 mol/L = 0.0949 mols (÷3->0.0316)

For Na₃PO₄: 100.00 ml x 1 L / 1000 ml x 0.955 mol / L = 0.0955 mols (÷2->0.0478)

Since 0.0316 is less than 0.0478, Ca(NO₃)₂ is limiting and will determine the amount of product formed.

Now use the moles of Ca(NO₃)₂ to find the grams of Ca₃(PO₄)₂ formed:

0.0949 mols Ca(NO₃)₂ x 1 mol Ca₃(PO₄)₂ / 3 mols Ca(NO₃)₂ x 310.18 g /mol Ca₃(PO₄)₂ = 9.812 g Ca₃(PO₄)₂

Percent yield = actual yield / theoretical yield (x100%)

actual yield = 8.630 g (given in the problem)

theoretical yield = 9.812 g (calculated above)

% yield = 8.630 g / 9.812 g (x100%) = 88.0% yield