How many g NO gas can be produced from 120 mL of 2.80 M HNO3 and excess Cu?
3Cu(s) + 8HNO3(aq) ---> 3Cu (NO3)2(aq) + 4H2O(l) + 2NO(g)
change 120 ml to l
.120 L
then divide M by L to get moles
2.80 M/.120 L
then take this and multiply it by the atomic weight or mass of NO
To calculate the amount of gas produced in this reaction, we need to use the stoichiometry of the balanced chemical equation. The coefficients in front of the reactants and products tell us the ratio of moles.
In this reaction, the ratio between Cu and NO is 3:2. It means that for every 3 moles of Cu, we have 2 moles of NO.
First, let's calculate the amount of Cu by using the given volume and concentration of HNO3.
1. Convert the volume of HNO3 to moles:
The molarity (M) of a solution represents the number of moles of the solute per liter of solution. Therefore, we can convert the volume (in mL) of solution to moles.
Given:
Volume of HNO3 = 120 mL
Molarity of HNO3 = 2.80 M
Convert the volume to liters:
120 mL = 120/1000 L = 0.120 L
Calculate the moles of HNO3:
Moles of HNO3 = Molarity * Volume
= 2.80 mol/L * 0.120 L
= 0.336 mol
2. Use the stoichiometry of the balanced equation to find the moles of Cu:
From the balanced equation, the ratio between Cu and HNO3 is 3:8. It means that for every 8 moles of HNO3, we have 3 moles of Cu.
Moles of Cu = (Moles of HNO3 * 3) / 8
= (0.336 mol * 3) / 8
= 0.126 mol of Cu
Now, we can use the stoichiometry again to find the moles of NO produced.
3. Calculate the moles of NO:
From the balanced equation, the ratio between Cu and NO is 3:2. It means that for every 3 moles of Cu, we have 2 moles of NO.
Moles of NO = (Moles of Cu * 2) / 3
= (0.126 mol * 2) / 3
= 0.084 mol of NO
Therefore, 0.084 moles of NO gas can be produced from 120 mL of 2.80 M HNO3 and excess Cu.