Im doing review for chem 20 finals and i dont remember how to do this :/ heeeeellp please??
How many ML of 2.50 mol/L HNO3 is required to dissolve an old copper penny with a mass of 3.94g according to the following unbalanced equation?
__Cu + __ HNO3 ------> Cu(NO3)2 + ___ NO + ___H20
How many ML of 2.50 mol/L HNO3 is required to dissolve an old copper penny with a mass of 3.94g according to the following unbalanced equation?
__Cu + __ HNO3 ------> Cu(NO3)2 + ___ NO + ___H20
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(3.94g Cu ) / (63.546 grams per mole) = .062 moles of Copper
(.062 mol Cu) x ( 8 / 3 molar ratio of Cu to HNO3 in the balanced equation ) = .165 moles of HNO3 required for the reaction.
( .165 moles HNO3 ) / (2.50 mol/L HNO3) = .0660 Liters of HNO3 solution required, or 66.0 mL.
3 Cu + 8 HNO3 = 3 Cu(NO3)2 + 2 NO + 4 H2O
(.062 mol Cu) x ( 8 / 3 molar ratio of Cu to HNO3 in the balanced equation ) = .165 moles of HNO3 required for the reaction.
( .165 moles HNO3 ) / (2.50 mol/L HNO3) = .0660 Liters of HNO3 solution required, or 66.0 mL.
3 Cu + 8 HNO3 = 3 Cu(NO3)2 + 2 NO + 4 H2O