A student carries out the precipitation reaction shown below, starting with 0.030 moles of calcium nitrate. The final mass of the precipitate is 2.9 g. Answer the questions below to determine the percent yield. 3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq) 1. a. Which product is the precipitate? b. How many moles of the precipitate would one expect to be produced from 0.030 moles of calcium nitrate? c. How many grams of solid do you expect to be produced? d. What is the percent yield?

a. Ca₃(PO₄)₂.

b. 0.010 moles of Ca₃(PO₄)₂ can we expect to be produced

c. 3.1g of Ca₃(PO₄)₂

d. Percent yield = 93.5%

Explanation:

a. Based on the reaction:

3Ca(NO₃)₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6NaNO₃(aq)

3 moles of calcium nitrate reacts with 2 moles of sodium phosphate producieng 1 mole of calcium phosphate.

As you can see, Ca₃(PO₄)₂ is a solid product -(s)-, that means when the reaction occurs the precipitate produced is the solid,

b. As 3 moles of calcium nitrate produce 1 mole of calcium phosphate and there are 0.030 moles of calcium nitrate

0.030 moles Ca(NO₃)₂ × (1 mol Ca₃(PO₄)₂ / 3 moles Ca(NO₃)₂) =

c. As molar mass of Ca₃(PO₄)₂ is 310.18g/mol, the mass of 0.010 moles (The expected mass) is;

0.010 moles Ca₃(PO₄)₂ × (310.18g / mol) =

d. The percent yield is defined as 100 times the ratio between the obtained yield (That is 2.9g of precipitate, Ca₃(PO₄)₂) and the expected yield, 3.1g of Ca₃(PO₄)₂:

i need mutiple choices

the answer is na (sodium)