How many moles of Al can be produced from 10.87 g of Ag?

Al (No3) 3(s) + 3Ag -> Al + 3AgNo3

0.91g of Al

Explanation:

Al(NO3)3(aq) + 3Ag(s) -> Al + 3Ag(NO3)

We check and we see that the given reaction equation is balanced

Now, from the equation, 3 moles of Ag reacts to produce 1 mole of Al

Mole ratio is 3:1

Number of moles = mass/molar mass

Molar mass of Ag ~ 108g/mol

3 moles of Ag = m/108g/mol

m = 3 moles of Ag × 108g/mol

m= 324g

1 mole of Al = 27g

Therefore, 324g of Ag produces 27g of Al,

10.87g of Ag will produce [(10.87g×27g)/324g] of Al

= 0.91g of Al

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

you need to have a picture of the atom that way we can answer it