Now, my problem with this question is the break-down, I don't know where to start.
Calculate the volume of H (g), measured at 21.9°C and 769 Torr required to react with 36.3 L CO(g), also measured at 21.9°C
and 769 Torr. Assume the gases react as follows.
3CO(g) + 7H2 (g) → C3H8 (g) + 3H2O(l)
Answer in 3 sig. digits.
Someone please help me, they're just practice questions but I need some help
Hydrogen peroxide, H O , is used to disinfect contact lenses. What volume of O , in mL, of O (g) at 36°C and 741 mmHg can be
liberated from 50.0 mL of an aqueous solution containing 3.00% H O by mass? The density of the aqueous solution of H O is
1.01 g/mL.
2H2O2( aq ) → 2H2O(l) + O2(g)
Enter your answer with 3 significant digits.
Enter scientific notation as 1.23E4
this is assignment question, go practice ur mom
To calculate the volume of H₂(g) that is required to react, we need to use the stoichiometry of the balanced chemical equation provided. In this case, we have the equation:
3CO(g) + 7H₂(g) → C₃H₈(g) + 3H₂O(l)
We know the volume of CO(g) is 36.3 L. However, we need to find the volume of H₂(g) required to react. To do this, we will use the concept of molar ratios.
Step 1: Convert the volume of CO(g) to moles
Use the ideal gas law to convert the volume of CO(g) to moles.
PV = nRT
Given:
P = 769 Torr (converted to atm by dividing by 760)
V = 36.3 L
T = 21.9°C (converted to Kelvin by adding 273.15)
R = 0.0821 L·atm/(K·mol)
Calculations:
P = 769 Torr / 760 Torr/atm = 1.012 atm
T = 21.9°C + 273.15 = 295.05 K
Using PV = nRT:
n = PV / RT
n = (1.012 atm) * (36.3 L) / (0.0821 L·atm/(K·mol) * 295.05 K)
n ≈ 1.722 moles CO
Step 2: Use the stoichiometry to find the moles of H₂(g)
From the balanced equation, we see that 7 moles of H₂(g) reacts with 3 moles of CO(g). Therefore, we can set up a ratio:
(7 moles H₂ / 3 moles of CO) = (x moles H₂ / 1.722 moles CO)
Calculations:
(7/3) = (x / 1.722)
x = (7/3) * 1.722
x ≈ 4.037 moles H₂
Step 3: Convert the moles of H₂(g) to volume
Now, we can use the ideal gas law again to find the volume of H₂(g).
Given:
P = 769 Torr (converted to atm by dividing by 760)
T = 21.9°C (converted to Kelvin by adding 273.15)
R = 0.0821 L·atm/(K·mol)
n = 4.037 moles H₂
Calculations:
P = 769 Torr / 760 Torr/atm = 1.012 atm
T = 21.9°C + 273.15 = 295.05 K
Using PV = nRT:
V = (nRT) / P
V = (4.037 moles) * (0.0821 L·atm/(K·mol)) * (295.05 K) / (1.012 atm)
V ≈ 94.53 L
So, the volume of H₂(g) required to react is approximately 94.53 L (rounded to 3 significant digits).