How many moles of cu are needed to react with 5.8 moles of agno3? cu + 2 agno3 → cu(no3)2 + 2 ag

moles of Cu° needed =  ½ (5.8 moles) = 2.9 moles Cu° needed.

Explanation:

Cu + 2AgNO₃  =>  Cu(NO₃)₂ + 2Ag

? moles Cu + 5.8 moles AgNO₃ =>    Cu(NO₃)₂ + 2Ag

Note coefficient of Cu° vs coefficient of AgNO₃

=> 1 mole Cu° < 2 moles AgNO₃ ...

=> Since moles of Cu° are smaller than moles of AgNO₃ in the given equation, the moles of Cu° needed to react with 5.8 moles AgNO₃ will be smaller than the 5.8 by the ratio of coefficients that will make 5.8 smaller. That is ...

moles of Cu° needed =  ½ (5.8 moles) = 2.9 moles Cu° needed.

Note: Using 2/1(5.8) will make value greater than 5.8 and incorrect.

One can also set up a ratio relationship as follows...

1 mole Cu° <=> 2 moles AgNO₃ (fm equation)

? mole Cu° <=> 5.8 moles AgNO₃ (problem)

=> (1 mole Cu°)/X=(2 mole AgNO₃)/(5.8 moles AgNO₃)

=> X = (1 mole Cu°)(5.8 mole AgNO₃)/2 mole AgNO₃

        = ½(5.8) mole Cu° = 2.9 mole Cu°

b) mutualism

hope this : )

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