If 12.4 moles of Cu and 48.2 moles of HNO3 are allowed to react, how many moles of excess reactant will remain if the reaction goes to completion? 3 Cu + 8 HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

Excess reactant amount 15.1 moles

Explanation:

3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO +4 H₂O

Using mole ratio method we find limiting reagent and excess reagent

For,

Since, Cu is limiting reagent and HNO₃ excess reagent

According to reaction,

              3 moles of Cu react with  8 moles of HNO₃

             12.4 moles of Cu react with = 33.06 moles of HNO₃

So remaining amount of excess   HNO₃  = 48.2 - 33.06

                                                                    = 15.1 moles

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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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.