How Much Gas is Produced from 8.8g of N2O Decomposition at STP?

[chawki]

Homework Statement

Dinitrogen monoxide (nitrous oxide) N₂O decomposes producing nitrogen and oxygen gases.
The atomic masses are N: 14.0 and O: 16.0
The molar volume of an ideal gas at STP is 22.4 dm3/mol

Homework Equations

Calculate the total volume of produced gases at STP, when 8.8 grams of dinitrogen monoxide decomposes?

The Attempt at a Solution

V_{total des gaz}=V_N+V_O

N₂O -------> 2N + O
8.8g---------m_N
44g/mol------14g/mol

m_N=(8.8*14)/44=2.8g
n_N=2.8/14=0.5mol.

22.4=V_N/n_N
V_N=22.4*n_N
V_N=22.4*0.5=11.2L?

 

[dextercioby]

2,8/14 =/= 0.5. Make the division again.

OK, what's V_O equal with ? Then just add the results.

 

[chawki]

It shouldn't be like this ?
N₂O -------> 2N + O
8.8g---------m_N
44g----------2*14g

then m_N=5.6g.
n_N=m_N/M_N=5.6/14=0.4mol

22.4=V_N/n_N
V_N=22.4*0.4=8.96L

Please tell me if this is correct, I'm confused if we should put 2*14 in this method or not. and then when calculating n, i used only 14.

if it's correct, we do same for O and then we add the volumes to each other to find the total volumes of gases.

 

[dextercioby]

The correct reaction should be of course

$$2\mbox{N}_{2}\mbox{O} \longrightarrow 2\mbox{N}_{2} + \mbox{O}_{2}$$

Apologies, I didn't check your equation for correction and jumped directly into the numbers.

Please, redo the calculations.

Thanks

 

[chawki]

2N₂O-------> 2N₂+O₂
8.8g----------m_N
2*44g--------2*28

m_N=(2*28*8.8)/88=5.6g

n_N=m_N/M_N
nN=5.6/14 (im not sure if we should divide by 14 or by (2*14))
n_N=0.4mol

22.4=V_N/n_N
V_N=22.4*0.4
V_N=8.96L

 

[dextercioby]

No, no. Because of the molecule being diatomic, the molar mass of nitrogen is 2x14 = 28 grams/mol. So it's 0.2 moles of nitrogen = 4.48 liters of nitrogen normal cond of temp and pressure.

 

[chawki]

I don't get it, i just finished editing my previous solution, please check it!

 

[dextercioby]

2N₂O-------> 2N₂+O₂
8.8g----------m_N
2*44g--------2*28

m_N=(2*28*8.8)/88=5.6g

n_N=m_N/M_N
nN=5.6/14 (im not sure if we should divide by 14 or by (2*14))
n_N=0.4mol

22.4=V_N/n_N
V_N=22.4*0.4
V_N=8.96L

The issue is in the bolded part. The number of moles of nitrogen is 5.6 grams/28 grams/mol = 0.2 mols.

So the volume which corresponds is 0.2 x 22.4 = 4.48l

 

[chawki]

ok so for O₂

2N₂O----------->2N₂+O₂
8.8g-----------------m_O2
2*44g---------------32g

m_O2= 3.2g
n_O2=m_O2/M_O2=3.2/32=0.1 mol of O₂

22.4=V_O2/n_O2
V_O2=2.24L

Vtotal=4.48+2.24=6.72L

should we write m_N or m_N2
same thing for O₂

 

[chawki]

No, no. Because of the molecule being diatomic, the molar mass of nitrogen is 2x14 = 28 grams/mol. So it's 0.2 moles of nitrogen = 4.48 liters of nitrogen normal cond of temp and pressure.

The 2*14 comes from N₂ or from 2N, i guess it's from N₂!

 

[dextercioby]

You need to check your arithmetics. The numbers in post #9 are wrong.

 

[chawki]

Yes it was a terrible mistake, please check the post #9

 

[dextercioby]

It looks ok now.

 

[chawki]

Thank you!