What method would you use to separate an aqueous solution of lead(ii) and copper(ii) ions
1 answer:
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Answer :
(A) The balanced molecular equation will be:
(B) The balanced molecular equation will be:
(C) The balanced molecular equation will be:
(D) The balanced molecular equation will be:
Explanation :
Molecular equation : It is defined as a balanced chemical equation where the ionic compounds are expressed in the form of molecules rather than component of ions.
Precipitation reaction : It is defined as the reaction in which an insoluble salt formed when two aqueous solutions are combined.
The insoluble salt that settle down in the solution is known an precipitate.
Part A : potassium carbonate and lead(II) nitrate
The balanced molecular equation will be:
In this reaction, lead carbonate is an insoluble salt and potassium nitrate is a soluble solution.
Part B : lithium sulfate and lead(II) acetate
The balanced molecular equation will be:
In this reaction, lead sulfate is an insoluble salt and lithium acetate is a soluble solution.
Part C : copper(II) nitrate and sodium sulfide
The balanced molecular equation will be:
In this reaction, Cuprous sulfide is an insoluble salt and sodium nitrate is a soluble solution.
Part D : strontium nitrate and potassium iodide
The balanced molecular equation will be:
In this reaction, strontium iodide and potassium nitrate are soluble solution.
Answer:
2.24 L of hydrogen gas, measured at STP, are produced.
Explanation:
Given, Moles of magnesium metal, = 0.100 mol
Moles of hydrochloric acid, = 0.500 mol
According to the reaction shown below:-
1 mole of Mg reacts with 2 moles of
0.100 mol of Mg reacts with 2*0.100 mol of
Moles of must react = 0.200 mol
Available moles of = 0.500 moles
Limiting reagent is the one which is present in small amount. Thus, is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of on reaction forms 1 mole of
0.100 mole of on reaction forms 0.100 mole of
Mole of = 0.100 mol
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Volume = ?
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
1 atm × V L = 0.100 × 0.0821 L.atm/K.mol × 273.15 K
<u>⇒V = 2.24 L</u>
2.24 L of hydrogen gas, measured at STP, are produced.