2 Au (CN)2 (aq) + Zn(s)
2Au(s) + Zn (CN)4 (ag)
1. How many grams of Zn are needed to react with 0.11 moles of Au (CN)2?
2. How many grams of Au can form from 0.11 mole of Au (CN)2?
Answers
Answer:
456 grams is required each during the process
1. If you write the reaction equation:
Zn + Au (CN)2 = Zn (CN)2 + Au
According to the stoichiometric coefficients, 1 equivalent of zinc requires 2 equivalents of cyanide ion.
As a result, in order for the reaction to be complete, all of the cyanide ions must be used up, i.e. all of the equivalents must react with zinc.
As a result, 0.11 moles of zinc would be required to complete the reaction (mole concept)
As a result, total mass = number of moles multiplied by molar mass
=0.11*65.4
= 7.194 grams
2. Apply the PRINCIPLE OF ATOM CONSERVATION (POAC) - It states that "atoms remain conserved during a chemical reaction."
According to the above question, all Au comes from Au(CN)2.
As a result, we apply POAC to 'Au' as follows:
The number of Au atoms in Au (CN)
2 moles of gold (CN)
2 = the number of Au atoms moles of Au, take this as equation (1)
It is given, (n)moles of Au(CN)2 = 0.11
Au molar mass = 197 g/mole
Au moles = given mass (w)/Molar mass (M) = w/197
Using Equation (1), we get
⇒ 1× 0.11 = 1 × w/197
⇒ w= 21.67.
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