Answer:
[tex]m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2[/tex]
Explanation:
Hello!
In this case, since the chemical reaction between copper and nitric acid is:
[tex]2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2[/tex]
By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:
[tex]m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2[/tex]
However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:
[tex]4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2[/tex]
Best regards!
Amount of copper (II) nitrate would be produced is 2.25 gram
Given that;
Amount of copper metal = 0.80 g
Find:
Amount of copper (II) nitrate would be produced
Computation:
Cu + 4HNO₃ → Cu(NO₃) + 2NO₂ + 2H₂O
Number of moles = Given mass / Molecular mass
Number of moles = 0.80 / 63.5
Number of moles = 0.012 mol
So,
n = Given mass / Molecular mass
0.012 = Given mass / 63.5 + (63.5 + 48)
Given mass = 2.25 gram
Amount of copper (II) nitrate would be produced = 2.25 gram
Find more information about 'Molecular mass'.
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