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Fittoniya [83]
3 years ago
15

A reaction is performed in a lab whereby two solutions are mixed together. The products are a liquid and a solid precipitate. Wh

at procedures would facilitate measurement of actual yield of the solid
Chemistry
1 answer:
tigry1 [53]3 years ago
7 0

Answer:

filtration, drying, and weighing

Explanation:

<em>The procedures that would facilitate the measurement of the actual yield of the solid would be filtration of the precipitate, drying of the precipitate, and weighing of the precipitate respectively.</em>

The liquid/solid mixture resulting from the reaction can be separated by the process of filtration using a filter paper. The residue in the filter paper would be the solid while the filtrate would be the liquid portion of the reaction's product.

The residue can then be allowed to dry, and then weighed using a laboratory-grade weighing balance. The weight of the solid represents the actual yield of the solid.

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Read 2 more answers
Which compound is soluble in water PbS BaS Na2S Fe2S3
Dmitry [639]

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The answer to your question is Na₂S, if you need to choices the other one is BaS

Explanation:

As a general rule, all the compounds that have sulfur, are insoluble in water, but the are some exceptions.

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3 years ago
A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →
vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

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