Question

How many moles of lithium nitrate are theoretically produced if we start with 3.0 moles of Ca(NO3)2 and 3.0 moles of Li3PO4? Reaction: 3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2

Answer 1

6.0 mol LiNO₃

Step-by-step explanation:

• 3Ca(NO₃)₂ + 2Li₃PO₄ → 6LiNO₃ + Ca₃(PO₄)₂

First we need to determine which reactant is the limiting reactant:

We calculate with how many Ca(NO₃)₂ moles would 3.0 moles of Li₃PO₄ react, using the stoichiometric coefficients of the reaction:

• 3.0 mol Li₃PO₄ *
  `(3molCa(NO_3)_2)/(2molLi_3PO_4)` = 4.5 mol Ca(NO₃)₂

There are not as many Ca(NO₃)₂ moles, so Ca(NO₃)₂ is the limiting reactant.

Using the number of moles of the limiting reactant, we calculate how many moles of LiNO₃ would be formed:

• 3.0 mol Ca(NO₃)₂ *
  `(6molLiNO_3)/(3molCa(NO_3)_2)` = 6.0 mol LiNO₃