# Given the balanced equation: 3Cu(s) + 8HNO3(aq) ---> 3Cu(NO3)2(aq) + 2N0(g) + 4H2O(l) determine the mass of copper that would be required to produce 4.00 L of nitrogen monoxide at 0.990 Atm and...

Given the balanced equation: 3Cu(s) + 8HNO3(aq) ---> 3Cu(NO3)2(aq) + 2N0(g) + 4H2O(l) determine the mass of copper that would be required to produce 4.00 L of nitrogen monoxide at 0.990 Atm and 22 degrees Celsius.

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First we need to use the gas law to calculate the number of moles of nitrogen monoxide (NO) produced. Using the equation PV=nRT, first rearrange to solve for n (the number of moles of gas):

PV=nRT

n=(PV)/(RT)

n = ((0.990 atm)*(4.00 L)) / ((0.0821 L*atm/mole*K) * 295.15 K) = 0.163 moles NO

From the balanced chemical equation above, we can see that 3 moles of copper (Cu) are required to produce 2 moles of NO. Using this conversion factor below:

0.163 moles NO * (3 moles Cu / 2 moles NO) = 0.245 moles copper.

Now we can convert the number of moles of copper into grams to get the mass of copper required:

0.245 moles Cu * (63.55 g / mole) = 15.57 grams copper.

**So 15.57 grams of copper are required.**

**Sources:**

When you see a problem like this that gives you information about a gas (nitrogen monoxide in this case) while it asks you for other information in the reaction (copper), then you should try the ideal gas law:

**PV=nRT**

Remember,

- P = pressure
- V = volume
- n = number of moles
- R = Gas constant (0.0821 L*atm/mol*K)
- T = temperature in Kelvins

The problem gives you information about nitrogen monoxide.

P = 0.990 atm

V = 4.00 L

You already have R (.0821 L*atm/mol*K).

T = 295 K (you get 295 by adding 273 to your 22 degrees Celsius. Always add 273 to your value in degrees Celsius to get your temperature in K)

You're just missing the value n. So what you do is plug in all your known values into the Ideal Gaw Law equation and then solve for n by doing algebra.

(0.990)(4.00) = n(0.0821)(295)

3.96 = (24.2195)(n)

0.1635 = n

Remember, n stands for 0.1635 moles of NO in your reaction. Now we can find the amount of grams of Cu with this value by doing stoichiometry and dimensional analysis.

`?g Cu = 0.1635 mol NOxx(3 mol Cu)/(2 mol NO)xx(63.55 g Cu)/(1 mol Cu) = 15.585 g Cu`

Now if you're required to do sig figs in this problem, look at the values in the question. They say 0.990 atm and 4.00 L. These values have 3 sig figs so make your answer 3 sig figs.

Your answer should be 15.6 grams of copper!

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When dealing with gases, your best bet is to use the Ideal Gas Law!

`PV = nRT`

Start off by plugging in your known variables (volume, pressure, and temperature). Temperature must be in Kelvins for this equation to work. R is the gas constant. I like to look at the units and the only gas constant that has both L and atm is: 0.0821 L atm/K mol.

`(0.990 atm)(4.0 L) = n(0.0821)(22+273 K)`

Solve for the only unknown variable, which is n in this case.

`3.96 = 24.2195 n`

`n = 0.164 "mol"`

The moles you found was for NO, but you are looking for Cu. Therefore, your last step is stoichiometry.

`0.164 "mol" = (3 "mol")/(2 "mol") = (63.54 "g")/(1 "mol") = 15.63 "g"`

Thus, you will need about 15.6 grams of copper (with three sig figs!).