Answer :
Answer:
239.8 g
Explanation:
This is a problem which involves stoichiometry calculations.
Every time we are to make a calculation based on a chemical reaction we do need to have it balanced:
SrCl₂ + Li₃PO₄ ⇒ LiCl + Sr₃(PO₄)₂ ( Double decomposition reaction )
We need two phophate in the reactants and that will give us 6 Li we balance in the products and the rest of the coefficients we will follow:
3 SrCl₂ +2 Li₃PO₄ ⇒ 6 LiCl + Sr₃(PO₄)₂
Now we can compute the moles of Li₃PO₄ that are going to react with the excess SrCl₂ .
From there we will calculate the moles of LiCl formed, and hence its mass by multiplying by its molar mass.
Molar Mass Li₃PO₄ = 115.79 g/mol
moles Li₃PO₄ = 218.3 g / 115.79 g/mol = 1.885 mol
moles LiCl formed from the stoichiometry of the reaction:
1.885 mol Li₃PO₄ x ( 6 mol LiCl / 2 mol Li₃PO₄ ) = 5.656 mol Li₃PO₄
molar mass LiCl = 42.39 g/mol
mass LiCl formed:
5.656 mol x 42.39 g/mol = 239.8 g
( rounded to 4 significant figures since we had 4 significant figures in the mass of 218.3 g of Li₃PO₄ )
The mass (in grams) of lithium chloride, LiCl formed is 239.9 g
Balanced equation for the reaction
3SrCl₂ + 2Li₃PO₄ —> Sr₃(PO₄)₂ + 6LiCl
Determination of the mass of Li₃PO₄ that reacted and the mass of LiCl produced
Molar mass of Li₃PO₄ = (3×7) + 31 + (16×4) = 116 g/mol
Mass of Li₃PO₄ from the balanced equation = 2 × 116 = 232 g
Molar mass of LiCl = 7 + 35.5 = 42.5 g/mol
Mass of LiCl from the balanced equation = 6 × 42.5 = 255 g
SUMMARY
From the balanced equation above,
232 g of Li₃PO₄ reacted to produce 255 g of LiCl
Determination of the mass of LiCl produced
From the balanced equation above,
232 g of Li₃PO₄ reacted to produce 255 g of LiCl
Therefore,
218.3 g of Li₃PO₄ will react to produce = (218.3 × 255) / 232 = 239.9 g of LiCl
Thus, 239.9 g of LiCl were obtained from the reaction.
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