1. How many liters of carbon are produced at STP when 295g of methane gas (C3H8) is reacted with excess oxygen? The other products is water vapor. Write out a balanced equation.
My answer:
C3H8+O—-> CO3+OH8?
your balanced equation stinks. Products are CO2 and H2O (water vapor)
C3H8 + 5*O2 = 3*CO2 + 4*H2O.
is it methane (CH4) or propane (C3H8) though?? eh, screw it, let's work with propane
1 mole of propane = 44 g - 3 moles of CO2 = 3*44 g
295 g of propane - x g of CO2
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x = (3*44*295)/44 g of CO2 is produces
let me now pretend CO2 is a perfect gas, in which every one of them has a volume of like 22,4 litres per mole
n(CO2) = x g/(44 g/mol) = x/44 moles
V(CO2) = 22,4 L/mol * n(CO2) mole = 22,4*n(CO2) L END
Is that all that I have to do because I feel as if there’s more for some reason. But if there isn’t can you show me how I should make my balanced equation look?
God can not give you the answer but He can guide you.
did you get an answer?
To balance the equation and answer the question, we first need to correctly write out the balanced equation for the reaction between methane (CH4) and oxygen (O2) to form carbon dioxide (CO2) and water (H2O).
The balanced equation for this reaction is:
CH4 + 2O2 → CO2 + 2H2O
Now, to find the number of liters of carbon dioxide produced, we need to convert the given mass of methane to moles and then use the mole-to-mole ratio from the balanced equation to determine the moles of carbon dioxide formed. Finally, we can convert the moles of carbon dioxide to liters using the ideal gas law.
Step 1: Convert the mass of methane to moles.
The molar mass of methane (CH4) is:
(1 atom of carbon x 12.01 g/mol) + (4 atoms of hydrogen x 1.01 g/mol) = 16.04 g/mol
295 g CH4 x (1 mol/16.04 g) ≈ 18.38 mol CH4
Step 2: Use the mole-to-mole ratio to determine moles of carbon dioxide.
From the balanced equation, we can see that 1 mol of CH4 produces 1 mol of CO2.
So, 18.38 mol CH4 will produce 18.38 mol CO2.
Step 3: Convert moles of carbon dioxide to liters.
Using the ideal gas law, PV = nRT, we can rearrange it to V = nRT/P,
where V represents the volume in liters, n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, and P is the pressure.
At STP (standard temperature and pressure), T = 273K and P = 1 atm.
Also, the ideal gas constant (R) is approximately 0.0821 L·atm/(mol·K).
V = (18.38 mol CO2) x (0.0821 L·atm/(mol·K)) x (273 K)/(1 atm)
V ≈ 429.35 L
Therefore, approximately 429.35 liters of carbon dioxide are produced at STP when 295g of methane gas is reacted with excess oxygen.