C6H12O6(s) + 9 O2(g) --------- 6 CO2(g) + 6 H2O(g) What volume of oxygen gas measured at 27oC and 0.987 atm is needed to react with 8 g of C6H12O6? (C-12.01, H-1.008, O-16.00).
First, your equation isn't balanced. Balance it.
mols C6H12O6 = grams/molar mass = ?
Using the coefficients in the balanced equation convert mols C6H12O6 to mols O2.
Then use PV = nRT and convert mols O2 to volume in L at the conditions listed.
To find the volume of oxygen gas needed to react with 8 g of C6H12O6, we need to use the given balanced equation and the ideal gas law.
First, let's calculate the moles of C6H12O6 using its molar mass:
8 g C6H12O6 x (1 mol C6H12O6 / 180.18 g C6H12O6) = 0.0444 mol C6H12O6
From the balanced equation, we can see that the stoichiometric ratio between C6H12O6 and O2 is 1:9. This means that for every 1 mole of C6H12O6, we need 9 moles of O2.
So, the number of moles of O2 required would be:
0.0444 mol C6H12O6 x (9 mol O2 / 1 mol C6H12O6) = 0.3996 mol O2
Now, we can use the ideal gas law to find the volume of the oxygen gas:
PV = nRT
Where:
P = pressure = 0.987 atm
V = volume (unknown)
n = number of moles of O2 = 0.3996 mol
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature in Kelvin = (27 + 273) K = 300 K
Rearranging the formula, we have:
V = nRT / P
Substituting the values:
V = (0.3996 mol) x (0.0821 L·atm/(mol·K)) x (300 K) / (0.987 atm)
V ≈ 9.83 L
Therefore, approximately 9.83 liters of oxygen gas measured at 27°C and 0.987 atm are needed to react with 8 g of C6H12O6.