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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

Respuesta :

Answer:

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,

    [tex]Cu\frac{mole}{stoichiometry} \frac{12.4}{3}=4.13[/tex]

[tex]HNO_3{\frac{mole}{stoichiometry} } \frac{48.2}{8}=6.025[/tex]

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

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