how many moles involved?
1 mole is 22.4L at stp
You burn 6 grams of carbon in plenty of air:
C ( s) + O2 ( g) = CO2 ( g)
a What volume of gas will form (at rtp)?
b What volume of oxygen will be used up?
10 answers
a) mass=6g Ar=12
Mr=12g/mol
n=m÷Mr
n=6g÷12g/mol
n=0.5 mol
V=n*Vm =0.5molx24dm(cubed)/mol
=12dm(cubed)
Mr=12g/mol
n=m÷Mr
n=6g÷12g/mol
n=0.5 mol
V=n*Vm =0.5molx24dm(cubed)/mol
=12dm(cubed)
57
E=MC SQUARE
THE E=MC SQUARE COMES IN DR.STONE ANIME.BUT 12 DM3 IS THERE BECAUSE THERE IS ONLY 6 G OF CARBON . 6 G OF CARBON HAS THE VOLUME OF 12DM3 WHICH IS 24DM3 DIVIDED BY TWO.SO DONT WORRY IF YOU HAVE ANY DOUBT YOU CAN CALL ME.9976852351
THE E=MC SQUARE COMES IN DR.STONE ANIME.BUT 12 DM3 IS THERE BECAUSE THERE IS ONLY 6 G OF CARBON . 6 G OF CARBON HAS THE VOLUME OF 12DM3 WHICH IS 24DM3 DIVIDED BY TWO.SO DONT WORRY IF YOU HAVE ANY DOUBT YOU CAN CALL ME.9976852351
Please Label which one is A and B
Hi what is a and what is b I don't know.
Go clean the dishes
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a) 12grams= 1 mole
6 grams= 0.5 mole
0.5 ×24= 12dm³
Answer=12dm³
6 grams= 0.5 mole
0.5 ×24= 12dm³
Answer=12dm³
b) From the balanced equation, we can see that for every 1 mole of carbon burned, 1 mole of oxygen is required. Therefore, since we have 0.5 moles of carbon, we also need 0.5 moles of oxygen.
Using the ideal gas law, we can calculate the volume of oxygen used:
PV = nRT
Assuming constant temperature and pressure (rtp):
V = nRT/P
V = (0.5 mol)(8.31 J/mol·K)(273 K)/(101.3 kPa)
V = 10.9 L or 10900 mL
Therefore, 10.9 L or 10900 mL of oxygen will be used up.
Using the ideal gas law, we can calculate the volume of oxygen used:
PV = nRT
Assuming constant temperature and pressure (rtp):
V = nRT/P
V = (0.5 mol)(8.31 J/mol·K)(273 K)/(101.3 kPa)
V = 10.9 L or 10900 mL
Therefore, 10.9 L or 10900 mL of oxygen will be used up.