Cu + 2 AgNO3 + Cu(NO3)2 + 2Ag If 7.28 moles of copper and 7.28 moles of silver nitrate are available to
react, what is the theoretical yield of silver produced?

7.28 moles Ag°

Explanation:

              Cu°          +    2 AgNO₃  => Cu(NO₃)₂ + 2Ag°

Given 7.28 moles     7.28 moles

To determine limiting reactant, divide the mole values by the respective coefficient of balanced equation. The resulting smallest value is the limiting reactant. Note: this is a short cut method for determining limiting reactant only. Once the limiting reactant is determined one must use the given mole values of the limiting reactant to solve problem. That is ...

Limiting reactant determination:

              Cu°          +    2 AgNO₃  => Cu(NO₃)₂ + 2Ag°

Cu: 7.28 / 1  = 7.28

AgNO₃ : 7.28 / 2 = 3.64 => Limiting Reactant is AgNO₃

Solving Problem depends on AgNO₃; Cu will be in excess.

Since coefficients of AgNO₃ & Ag° are equal, then the moles AgNO₃ used equals moles Ag° produced and is therefore 7.28 moles Ag°.

the answer is the first option

explanation:

just googled it m8. not that hard

lol