Respuesta :
Answer:
[tex]3.37~g~Li_3PO_4[/tex]
Explanation:
The first step is to find the moles of [tex]Li^+[/tex], using the equation:
[tex]M=\frac{mol}{L}[/tex]
Where, [tex]500 mL=0.5 L[/tex]
Plug in the values into the equation:
[tex]0.175~M=\frac{mol}{0.5~L}[/tex]
Solving for mole:
[tex]mol=0.175~M*0.5~L[/tex]
[tex]mol=0.0875~mol~Li^+[/tex]
Now, the have to write the chemical equation to find the molar ratio between [tex]Li_3PO_4[/tex] and [tex]Li^+[/tex], so:
[tex]Li_3PO_4~->~3Li^+~+~(PO_4)^-3[/tex]
The molar ratio then is 1:3, using this molar ratio we can convert from moles of [tex]Li^+[/tex] to moles of [tex]Li_3PO_4[/tex], so:
[tex]mol=0.0875~mol~Li^+~\frac{1~mol~Li_3PO_4}{3~mol~Li^+} =0.0291~mol~Li_3PO_4[/tex]
Finally, we have to conver from moles of [tex]Li_3PO_4[/tex] to grams of [tex]Li_3PO_4[/tex] using the molar mass of [tex]Li_3PO_4[/tex] (115.79 g/mol), so:
[tex]0.0291~mol~Li_3PO_4\frac{115.79~gLi_3PO_4}{1~mol~Li_3PO_4}=3.37~g~Li_3PO_4[/tex]
The proportionate of the moles of the substances and the volume in litres of the substances gives the molarity(M).
3.38 g of [tex]\rm Li_{3}PO_{4}[/tex] prepare 500 ml of a solution.
How to calculate the mass?
The formula for calculating Molarity is:
[tex]\rm Molarity (M) = \dfrac{\rm moles\; (n)}{\;\rm Volume \;(V)}[/tex]
First, calculate the moles of [tex]\rm Li^{+}[/tex]:
Given,
- Molarity of [tex]\rm Li^{+}[/tex] = 0.175 M
- Volume (V) = 0.5 L
[tex]\begin{aligned}0.175 &= \dfrac{\rm moles}{0.5 \;\rm L}\\\\\rm n &= 0.175 \times 0.5\;\rm L\\\\&= 0.0875 \;\rm moles\end{aligned}[/tex]
The chemical reaction can be shown as,
[tex]\rm Li_{3}PO_{4} \rightarrow 3Li^{+} + (PO_{4})^{-3}[/tex]
From the reaction, it can be said that,
1 mole of [tex]\rm Li_{3}PO_{4}[/tex] = 3 moles of [tex]\rm 3Li^{+}[/tex]
X moles of [tex]\rm Li_{3}PO_{4}[/tex] = 0.0875 moles of [tex]\rm 3Li^{+}[/tex]
Solving further for X:
[tex]\begin{aligned}\rm X &= \dfrac{0.0875 \times 1 \;\rm mole}{3 \;\rm moles}\\\\\rm X &= 0.0291 \;\rm moles \end{aligned}[/tex]
Now, mass can be calculated as,
[tex]\begin{aligned}\rm Moles & = \dfrac{\rm mass}{\rm molar\; mass}\\\\\rm m &= 0.0291 \times 115.79\\\\&= 3.37 \;\rm g\end{aligned}[/tex]
Therefore, 3.37 gm of [tex]\rm Li_{3}PO_{4}[/tex] is needed.
Learn more about molarity here:
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