Answer:
Volume of the concentrated solution, which is needed is 103.30 mL
Explanation:
Let's apply the formula for dilutions to solve the problem
Conc. Molarity . Conc. volume = Dil. Molarity . Dil volume
12.1 M . Conc. volume = 2.5 M . 500 mL
Conc. volume = (2.5 M . 500 mL) / 12.1M
Conc. volume = 103.30 mL
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Answer:
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°.
