Answer to Question #104538 in General Chemistry for ReAnna Rodriguez

Question

what mass of copper II nitrate can be produced from a reaction of 3.72 grams of copper with 9.83 grams of silver nitrate, according to the following equation?

Answer 1

Cu+2AgNO₃ → 2Ag + Cu(NO₃)₂

n=m/M ; n - number of moles; m - mass of substance; M - molar mass of the substance;

n(Cu)= 3.72/64 = 0.058 moles;

M(AgNO₃)= 108+14+48=170;

n(AgNO₃) =9.83/170= 0.058 moles;

In order to know which of two reagents will determine the mass of copper II nitrate produced, we need to understand which of them is the excess reactant and the limiting reactant according to this reaction and the initial number of moles. The easiest way to do that is dividing their number of moles by respective coefficients:

0.058/1=0.058 - for Cu;

0.058/2=0.029 - for AgNO₃;

So, AgNO₃ is the limiting reactant. Now we can calculate the mass of copper II nitrate produced:

0.058/2=m(Cu(NO3)₂)/M(Cu(NO3)₂); M(Cu(NO3)₂)= 188;

m(Cu(NO3)₂)=0.058*188/2 = 5.452 grams;

Answer: The mass of copper II nitrate that can be produced is 5.452 grams.