Question

Consider the following balanced equation:

3Ca(NO₃)₂ + 2 Li3PO₄ --> 6LINO₃ + Ca₃(PO₄)₂
How many grams of calcium phosphate, Ca₃(PO₄)₂, are theoretically produced if we start with 1.75 moles of Ca(NO₃)₂ and 2.09 moles of Li₃PO₄?
a. 543 grams
b. 181 grams
c. 326 grams
d. 648 grams

Answer 1

To determine the number of grams of calcium phosphate, Ca₃(PO₄)₂, produced, we need to calculate the limiting reactant first. The correct answer is approximately 181 grams (option b).

Step-by-step explanation:

To determine the number of grams of calcium phosphate, Ca₃(PO₄)₂, produced, we need to calculate the limiting reactant first.

Converting the moles of Ca(NO₃)₂ to moles of Ca₃(PO₄)₂:

1.75 moles Ca(NO₃)₂ x (1 mole Ca₃(PO₄)₂ / 3 moles Ca(NO₃)₂) = 0.583 moles Ca₃(PO₄)₂

Converting the moles of Li₃PO₄ to moles of Ca₃(PO₄)₂:

2.09 moles Li₃PO₄ x (1 mole Ca₃(PO₄)₂ / 2 moles Li₃PO₄) = 1.045 moles Ca₃(PO₄)₂

Since the stoichiometry of the balanced equation shows that 2 moles of Ca₃(PO₄)₂ are produced for every 2 moles of Li₃PO₄, the limiting reactant is Ca(NO₃)₂.

Therefore, the number of moles of Ca₃(PO₄)₂ produced is 0.583 moles.

To convert moles to grams, we need to use the molar mass of Ca₃(PO₄)₂, which is approximately 310.18 g/mol:

0.583 moles Ca₃(PO₄)₂ x (310.18 g / 1 mole) = 180.74 grams

Therefore, the correct answer is approximately 181 grams (option b).