Question

part a
how many moles of lithium nitrate are theoretically produced if we start with 3.4 moles of ca(no₃)₂ and 2.4 moles of li₃po₄?
reaction: 3ca(no₃)₂ + 2li₃po₄ → 6lino₃ + ca₃(po₄)₂
options: 1.1, 7.2, 6.8, 1.2, not enough information
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Explanation:

Step1: Determine Limiting Reactant

From the reaction \( 3\text{Ca(NO}_3\text{)}_2 + 2\text{Li}_3\text{PO}_4
ightarrow 6\text{LiNO}_3 + \text{Ca}_3(\text{PO}_4)_2 \), the mole ratio of \( \text{Ca(NO}_3\text{)}_2 \) to \( \text{Li}_3\text{PO}_4 \) is \( 3:2 \).
For \( 3.4 \) mol \( \text{Ca(NO}_3\text{)}_2 \), moles of \( \text{Li}_3\text{PO}_4 \) required: \( \frac{2}{3} \times 3.4 \approx 2.27 \) mol. We have \( 2.4 \) mol \( \text{Li}_3\text{PO}_4 \), so \( \text{Ca(NO}_3\text{)}_2 \) is limiting.

Step2: Calculate Moles of \( \text{LiNO}_3 \)

Mole ratio of \( \text{Ca(NO}_3\text{)}_2 \) to \( \text{LiNO}_3 \) is \( 3:6 = 1:2 \).
Moles of \( \text{LiNO}_3 = 2 \times 3.4 = 6.8 \) mol.

Answer:

6.8 (corresponding to the option "6.8")