We mix 112 grams of oxygen gas with 118 grams of argon gas in a volume of 1752 mL at 74 degrees celsius. What will be the final pressure of the gas mixture?
First, I converted the grams into moles.
112g O2 * (1mol O2/ 32g O2) = 3.5 moles O2
118g Ar * (1mol Ar/ 40g Ar) = 2.95 moles O2
Then, I added the moles to get 6.45 moles ArO2
I used, P=nrT/V
=(6.45mol)(0.08206Latm/mol)(74/273)/1.752L
=104.83 atm
Is this right? Somehow, I am not getting correct sigfigs.
I think you have a typo in your post. T is 74+ 273 and you show it as 74/273; however, you must have used 347 because my answer comes out 104.83 atm also. I would round that to 105. You have 3 s.f. in the 112 and 118. Is that 74.0 C?
Your calculations appear to be correct. However, there is a mistake in your final answer regarding significant figures.
To determine the correct number of significant figures, you need to consider the given values with the fewest significant figures. In this case, the volume of 1752 mL has four significant figures.
Using the given values, let's recalculate the final pressure with the appropriate number of significant figures:
P = (6.45 mol)(0.08206 L·atm/mol·K)(74 °C + 273) / 1.752 L
= (6.45)(0.08206)(347) / 1.752
= 106.0774 atm
Since the volume has four significant figures, we should round the final answer to four significant figures as well:
Final pressure = 106.1 atm
Therefore, the correct final answer, considering significant figures, is 106.1 atm.
Your calculation is almost correct, but you made a small error in determining the total number of moles. Let's correct it and recalculate.
To find the total number of moles of the gas mixture, you need to add the moles of oxygen gas (O2) and argon gas (Ar) separately.
Moles of O2 = 112g O2 * (1 mol O2 / 32g O2) = 3.5 moles O2
Moles of Ar = 118g Ar * (1 mol Ar / 40g Ar) = 2.95 moles Ar
Total moles of gas mixture = 3.5 moles O2 + 2.95 moles Ar = 6.45 moles ArO2 (which is correct)
Now, let's substitute the correct values into the ideal gas law equation:
P = (n * R * T) / V
P = (6.45 mol) * (0.08206 Latm/molK) * (74 + 273) K / 1.752 L
P = 14.295 atm
The final pressure of the gas mixture will be approximately 14.3 atm.
Regarding significant figures, it is important to carry out the calculations with all available significant figures and then round the final answer to the appropriate number of significant figures. In this case, since the given data has three significant figures, the final answer should also be rounded to three significant figures, resulting in 14.3 atm.