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 2 years ago
A .411g sample of powder copper mixed with an inert, soluble additive was fully consumed by 23.4 mL of .602M nitric acid, producing copper (II) nitrate, water and nitric oxide (NO) gas. What was the percent copper by mass in the sample? 3Cu + 8HNO3 > 3 Cu(NO3) + 2NO + 4 H2O
 2 years ago
A .411g sample of powder copper mixed with an inert, soluble additive was fully consumed by 23.4 mL of .602M nitric acid, producing copper (II) nitrate, water and nitric oxide (NO) gas. What was the percent copper by mass in the sample? 3Cu + 8HNO3 > 3 Cu(NO3) + 2NO + 4 H2O

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agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0You'll need to know the atomic mass of elemental copper, hydrogen, nitrogen, and oxygen (off the periodic table) first thing, you'll be just doing some simple addition here. This is the mass in grams for ONE mole of that element (with isotopes already taken into account). You already have the ratios of the reaction so you can find the theoretical yield. Where this could get tricky is if you need to take into account that the reaction isn't going to be 100% efficient as written in the real world (actual yield is less than theoretical, sometimes by a lot). And if you need to take into account as the nitric acid gets used up, it starts to slow don. In this case I think you have enough nitric acid to run this reaction to products completely and your limiting reagent is the solid copper. I think you might have written the question down wrong slightly, make sure it doesn't say "What is the mass of copper in the Copper Nitrate sample?" Because that would make a whole lot more sense for these classic types of problems. If that's the case all you're needing to do is find the quantity of products produced using the mole ratios shown in the equation, and then find out the mass for that and how much copper accounts for that. How were you introduced to this type of problem in class?

onmypagrind
 2 years ago
Best ResponseYou've already chosen the best response.0I wrote the question out as stated on the handout that we were given in class. I am still out in left field even with your explanation

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0Doing step #1 for you ;) Hyrdrogen \[\frac{1.00784 \ \text{g H}}{1\ \text{mol H}}\] Carbon \[\frac{12.011 \ \text{g C}}{1\ \text{mol C}}\] Nitrogen \[\frac{14.00674 \ \text{g H}}{1\ \text{mol H}}\] Oxygen \[\frac{15.9994 \ \text{g H}}{1\ \text{mol H}}\] Copper \[\frac{63.546 \ \text{g H}}{1\ \text{mol H}}\] Now how many moles of Copper did this reaction start out with?
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