Determine the minimum volume of 6.40M HNO3 required to dissolve a 2.40g sample of Cu.
2NO3-(aq)+8H+(aq)+3Cu(s)-->2NO(g)+4H2)(l)+3Cu2+(aq)
Convert 2.40 g Cu to moles. moles = grams/molar mass
Using the coefficients in the balanced equation, convert moles Cu to moles HNO3.
Molarity = moles/L. YOu know moles HNO3 and M HNO3, calculate L.
To determine the minimum volume of 6.40M HNO3 required to dissolve a 2.40g sample of Cu, we can use stoichiometry and the balanced chemical equation provided.
First, we need to determine the number of moles of Cu in the 2.40g sample. We can do this by using the molar mass of Cu:
Molar mass of Cu = 63.55 g/mol
Number of moles of Cu = mass of Cu / molar mass of Cu
= 2.40g / 63.55 g/mol
= 0.0378 mol
From the balanced chemical equation, we can see that the stoichiometric ratio between Cu and HNO3 is 3:8. This means that for every 3 moles of Cu, we need 8 moles of HNO3.
Therefore, we can calculate the number of moles of HNO3 required by setting up a ratio:
Number of moles of HNO3 = (number of moles of Cu) * (8 moles of HNO3 / 3 moles of Cu)
= 0.0378 mol * (8/3)
= 0.1008 mol
Now, we can calculate the minimum volume of 6.40M HNO3 required by using the equation:
Molarity (M) = moles / volume (L)
Rearranging the equation to solve for volume:
Volume (L) = moles / molarity (M)
Volume of 6.40M HNO3 required = 0.1008 mol / 6.40 mol/L
= 0.01575 L
= 15.75 mL
Therefore, the minimum volume of 6.40M HNO3 required to dissolve the 2.40g sample of Cu is 15.75 mL.
To determine the minimum volume of 6.40M HNO3 required to dissolve a 2.40g sample of Cu, we need to consider the stoichiometry of the reaction and use the concept of limiting reagent.
Let's start by writing down the balanced chemical equation:
2NO3-(aq) + 8H+(aq) + 3Cu(s) --> 2NO(g) + 4H2O(l) + 3Cu2+(aq)
From the equation, we can see that the molar ratio between Cu(s) and HNO3 is 3:8. This means that for every 3 moles of Cu(s), we need 8 moles of HNO3 to completely react.
First, we need to convert the mass of Cu(s) to moles. The molar mass of Cu is 63.55 g/mol:
2.40 g Cu × (1 mol Cu / 63.55 g Cu) = 0.0378 mol Cu
Since the molar ratio between Cu(s) and HNO3 is 3:8, we can set up a ratio to find the amount of HNO3 needed:
0.0378 mol Cu × (8 mol HNO3 / 3 mol Cu) = 0.1016 mol HNO3
Now, we can calculate the volume of 6.40M HNO3 needed using the molarity formula:
Volume (L) = moles of solute / molarity
Volume (L) = 0.1016 mol HNO3 / 6.40 mol/L = 0.0159 L
Finally, we can convert the volume from liters to milliliters (mL):
0.0159 L × 1000 mL/L = 15.9 mL
Therefore, the minimum volume of 6.40M HNO3 required to dissolve the 2.40g sample of Cu is 15.9 mL.