What volume of 0.100 M Na3PO4 is required to precipitate all the lead(ii) ions from 150.0 mL of 0.250 M Pb(NO3)2?
2 answers:
Volume of 0.100 M Na₃PO₄ is 0.25 L
<h3>Further explanation </h3>Stoichiometry in Chemistry learn about chemicals mainly emphasizes quantitative, such as the calculation of volume, mass, number, which is related to numbers, molecules, elements, etc.
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
In the reaction there are also manifestations of reagent substances namely gas (g), liquid (liquid / l), solid (solid / s) and solution (aqueous / aq).
The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.
- Mole
The mole itself is the number of particles contained in a substance amounting to 6.02.10²³
Mole can also be sought if the amount of substance mass and its molar mass is known
Reaction that happens :
2Na₃PO₄ + 3Pb(NO₃)₂ ⇒ 6NaNO₃ + 3Pb₃(PO₄)₂
Pb(NO₃)₂ ⇒ Pb²⁺ + 2NO₃⁻
mole Pb²⁺ = mole Pb(NO₃)₂
mole Pb²⁺ = 0.15 L(150 ml) . 0.25 M = 0.0375 mole
mole Na₃PO₄ : mole Pb(NO₃)₂ = 2:3
mole Na₃PO₄ = 2/3 . 0.0375
mole Na₃PO₄ = 0.025
So volume of 0.100 M Na₃PO₄ =
volume = mole : M
volume = 0.025 : 0.1 = 0.25 L
<h3>Learn more</h3>The mass of one mole of raindrops
moles of NaOH
moles of water you can produce
Keywords: mole, precipitate, Na₃PO₄, Pb(NO₃)₂, volume
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The question in incomplete, complete question is;
Determine the theoretical yield:
Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.
Answer:
The theoretical yield of sodium nitrate is 340 grams.
Explanation:
Moles of copper(II) nitrate =
Moles of sodium sulfide =
According to reaction, 1 mole of copper (II) nitrate reacts with 1 mole of sodium sulfide.
Then 2 moles of sodium sulfide will react with:
of copper (II) nitrate
As we can see from this sodium sulfide is present in limiting amount, so the amount of sodium nitrate will depend upon moles of sodium sulfide.
According to reaction, 1 mole of sodium sulfide gives 2 mole of sodium nitrate, then 2 mole of sodium sulfide will give:
sodium nitrate
Mass of 4 moles of sodium nitrate :
85 g/mol × 4 mol = 340 g
Theoretical yield of sodium nitrate = 340 g
The theoretical yield of sodium nitrate is 340 grams.
Answer : The correct option is, They are different units of concentration.
Explanation :
Molarity : It is defined as the number of moles of solute present in one liter of solution.
Formula of molarity :
Mass percent : It is defined as the mass of solute present in the mass of solution.
Formula of mass percent :
Hence, both are the different unit of concentration.