3cu(s)+8HNO3=3cu(No3)2+2NO+4H2O(1)
How many nitric oxide, NO are formed after mixing if we have 0.25mol cu with 0.7 mol HNO3
I have rewritten the equation to help understand it. I assume you want to know number of mols NO formed under these conditions.
3Cu(s) + 8HNO3 => 3Cu(NO3)2 + 2NO + 4H2O(l)
mols Cu = 0.25
mols HNO3 = 0.7
mols NO from Cu if exess of HNO3 =
0.25 x (2 mol NO/3 mols Cu) = 0.25 x 2/3 = 0,167
mols NO from HNO3 if Cu in excess =
0.7 x (2 mols NO/8 mols HNO3) = 0.175
In limiting reagent problems the SMALLER number always wins (you can't get more than the smallest) so Cu is limiting reagent and you will form 0.167 mols NO.
To determine the number of nitric oxide (NO) molecules formed after mixing, we need to use stoichiometry. The balanced equation for the reaction is:
3Cu(s) + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O
From the balanced equation, we can see that the stoichiometric coefficient of NO is 2. This means that for every 3 moles of copper (Cu) and 8 moles of nitric acid (HNO3) used in the reaction, we get 2 moles of nitric oxide (NO).
Given that we have 0.25 mol of Cu and 0.7 mol of HNO3, we need to figure out the limiting reactant (the reactant that determines the amount of product formed). To do this, we compare the stoichiometric ratios of the two reactants.
For Cu:
0.25 mol Cu x (8 mol HNO3 / 3 mol Cu) = 0.67 mol HNO3
For HNO3:
0.7 mol HNO3
Since the calculated amount for Cu (0.67 mol HNO3) is less than the amount given for HNO3 (0.7 mol HNO3), Cu is the limiting reactant.
Now, we can determine the amount of NO formed using the stoichiometric ratio:
0.25 mol Cu x (2 mol NO / 3 mol Cu) = 0.17 mol NO
Therefore, if we have 0.25 mol of Cu and 0.7 mol of HNO3, the amount of NO formed will be 0.17 mol.