Okay so I did this first question:
1. In a lab, 2.01g copper reacted with 25mL of 6.0M HNO3 solution according to the following equation cu+4HNO3 -> cu(NO3)2+2NO2 +2H2O. In this lab, copper was the limiting reactant. Show work to support this statement.
I answered it this way:
Moles Cu: 2.01x 1mol/63.55g= 0.03163 mol
Moles HNO3= concentration X volume; which is 0.025L (cuz I converted milliliters to liters) X 6.01 mol/L = 0.150 mol
4HNO3 and Cu = 1:4 ratio
So, 0.03163 mol cu X 4= 0.130 mol HNO3
So we have more than enough to use up all the copper
Limiting reactant= copper
But the second and third questions I do not get:
2. In the above reaction, what volume of HNO3 solution was left over?
Is 0.2 right? And what would the units be if it was?
3. In the above reaction, what mass of Cu(NO3)2 was produced theoretically?
Do I just look it up in the periodic table? Is that it? Or does it have something to do with the ratio?
PLEASE HELP! I have so much to do and this is confusing me so much!
1. In a lab, 2.01g copper reacted with 25mL of 6.0M HNO3 solution according to the following equation cu+4HNO3 -> cu(NO3)2+2NO2 +2H2O. In this lab, copper was the limiting reactant. Show work to support this statement.
I answered it this way:
Moles Cu: 2.01x 1mol/63.55g= 0.03163 mol
Moles HNO3= concentration X volume; which is 0.025L (cuz I converted milliliters to liters) X 6.01 mol/L = 0.150 mol
4HNO3 and Cu = 1:4 ratio
So, 0.03163 mol cu X 4= 0.130 mol HNO3
So we have more than enough to use up all the copper
Limiting reactant= copper
But the second and third questions I do not get:
2. In the above reaction, what volume of HNO3 solution was left over?
Is 0.2 right? And what would the units be if it was?
3. In the above reaction, what mass of Cu(NO3)2 was produced theoretically?
Do I just look it up in the periodic table? Is that it? Or does it have something to do with the ratio?
PLEASE HELP! I have so much to do and this is confusing me so much!
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2. Tricky question because the original 25 ml of solution will still be there, except as you calculated, .13 moles of the HNO3 was reacted, leaving .02 moles of HNO3 left unreacted swimming around in 25 ml of solution.
the new molarity of the HNO3 would now be .02 moles HNO3 over .025 Liters = .8 Molar HNO3
3. Mass of Cu(NO3)2 since there is one mole of Cu inside one mole of Cu(NO3)2, and you reacted .0316 moles of Cu, you should produce .0316 moles of Cu(NO3)2
so .0316 moles Cu(NO3)2 times 187 grams/mole = 5.9 grams of copper nitrate
the new molarity of the HNO3 would now be .02 moles HNO3 over .025 Liters = .8 Molar HNO3
3. Mass of Cu(NO3)2 since there is one mole of Cu inside one mole of Cu(NO3)2, and you reacted .0316 moles of Cu, you should produce .0316 moles of Cu(NO3)2
so .0316 moles Cu(NO3)2 times 187 grams/mole = 5.9 grams of copper nitrate
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