Respuesta :
Explanation:
Given parameters:
Number of moles of Cu = 0.696mol
Volume of HNO₃ = 146mL
Concentration of HNO₃ = 5.5M
Unknown:
Will Cu react completely = ?
Limiting reagent = ?
Remaining mass of the compound = ?
Solution:
The equation of the reaction is given as;
Cu + 4HNO₃ → Cu(NO₃)₂ + 2H₂O + 2NO₂
Let us find the number of moles of HNO₃ ;
Number of moles = concentration x volume
Number of moles = 5.5 x 146 x 10⁻³L = 0.803mol
From the balanced reaction equation;
1 mole of Cu reacted with 4 mole of HNO₃
0.696 mole of Cu will require 4 x 0.696 = 2.784mole of HNO₃
So, the limiting reactant is HNO₃ because it is in short supply.
Cu will completely react
Mass of the remaining compound we need;
mass = number of moles x molar mass
Number of moles = 2.784 - 0.803 = 1.981mole
So, mass = 1.981 x (1 + 14 + 3(16)) = 124.81g of HNO₃ is required to make the reaction complete.
The limiting reagent is HNO3 and the remaining compound in mass is 25.1 grams of Cu.
The reaction taking place in the given case is,
Based on the given information,
• The moles of Cu given is 0.696 moles.
• The volume of HNO3 given is 146 ml and the molarity of HNO3 is 5.5 M.
Now the moles of HNO3 can be determined as,
[tex]Moles = \frac{M1V1}{1000} \\Moles = \frac{5.5M*146 ml}{1000} \\Moles = 0.803 moles[/tex]
• In order to find the limiting reagent, there is a need to divide moles by their stoichiometric coefficient, that is,
Cu = 0.696/3 = 0.232 moles and HNO3 = 0.803/8 = 0.100 moles
• As the moles of HNO3 is present in less amount, therefore, HNO3 will be the limiting reagent.
Now the remaining copper left in the reaction will be,
[tex]= 0.696-\frac{3}{8} *0.803\\= 0.696-0.300\\= 0.396 moles[/tex]
Now the mass of remaining copper will be = Moles * Molecular mass of Cu
[tex]Mass = 0.396 moles * 63.5 g/mol\\Mass = 25.1 g[/tex]
Thus, the mass of remaining compound is 25.1 grams and the limiting reagent is HNO3.
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