To determine how much Lithium Phosphide (Li₃P) is required to fully react with 8.70 x 10⁻⁵ moles of Magnesium Chloride (MgCl₂), we need to refer to the balanced chemical equation for their reaction. The chemical equation for this reaction is:
[tex]\[ 3 \text{MgCl}_{2} + 2 \text{Li}_{3}\text{P} \rightarrow 6 \text{LiCl} + \text{Mg}_{3}\text{P}_{2} \][/tex]
From the balanced equation, we can see that:
- 3 moles of MgCl₂ react with 2 moles of Li₃P.
This means that:
- 1 mole of MgCl₂ reacts with [tex]\(\frac{2}{3}\)[/tex] moles of Li₃P.
Given:
- Amount of MgCl₂ = 8.70 x 10⁻⁵ moles
To find the moles of Li₃P needed, we can use the molar ratio derived from the balanced equation:
[tex]\[ \text{Moles of Li}_{3}\text{P} = \left( \frac{2}{3} \right) \times \text{Moles of MgCl}_{2} \][/tex]
Substituting the given moles of MgCl₂:
[tex]\[ \text{Moles of Li}_{3}\text{P} = \left( \frac{2}{3} \right) \times 8.70 \times 10^{-5} \][/tex]
[tex]\[ \text{Moles of Li}_{3}\text{P} = \frac{2 \times 8.70 \times 10^{-5}}{3} \][/tex]
[tex]\[ \text{Moles of Li}_{3}\text{P} = \frac{17.40 \times 10^{-5}}{3} \][/tex]
[tex]\[ \text{Moles of Li}_{3}\text{P} = 5.80 \times 10^{-5} \][/tex]
Therefore, the amount of Lithium Phosphide required to fully react with 8.70 x 10⁻⁵ moles of Magnesium Chloride is [tex]\(5.8 \times 10^{-5} \)[/tex] moles of Li₃P.
The correct answer is:
- c 5.8 x 10⁻⁵ mol Li₃P
