This is a stoichiometry equation.
Cu(s) + 4(HNO3)(aq) -> Cu(NO3)2(aq) + 2(NO2)(g)+2(H2O)(l)
If you use 6 moles of nitric acid 4(HNO3), how many moles of Copper (II) Nitrate Cu(No3)2 will be produced?
HNO3/CU(NO3)2 = 4 mols /1 mol
so
6/x = 4/1
4 x = 6
x = 1.5
To determine the number of moles of Copper (II) Nitrate (Cu(NO3)2) that will be produced when 6 moles of nitric acid (HNO3) react, we need to use the stoichiometric coefficients from the balanced equation.
According to the balanced equation:
1 mole of Copper (Cu) reacts with 4 moles of nitric acid (HNO3) to produce 1 mole of Copper (II) Nitrate (Cu(NO3)2).
Therefore, if we have 6 moles of nitric acid, we can set up a proportion to find the number of moles of Copper (II) Nitrate produced:
4 moles HNO3 / 1 mole Cu(NO3)2 = 6 moles HNO3 / x moles Cu(NO3)2
Cross-multiplying the equation, we have:
4x = 6
x = 6/4
x = 1.5 moles Cu(NO3)2
So, if you use 6 moles of nitric acid (HNO3), 1.5 moles of Copper (II) Nitrate (Cu(NO3)2) will be produced.