Balancing Equations and Stoichiometry

Question

You burn some acetylene, C2H2, according to the unbalanced reaction. C2H2 + O2 —> CO2 + H2O.

A) How many grams of acetylene would be required to produce 63 L of CO2 gas at 1000 C?

B) How many grams of water would this produce?

Answer 1

Once again, the problem has neglected to provide the pressure, so once again, we will assume it to be 1 atm.

First thing we must do is to balance the equation:

2C₂H₂ + 5O2 ==> 4CO₂ + 2H₂O .. balanced equation

Next, we will find the moles of CO2 equivalent to 63 L of CO₂ @ 1000ºC:

Use the ideal gas law, PV = nRT

P = pressure = 1 atm (assumed)

V = volume in L = 63 L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 1000C + 273 = 1273K

Solving for n, we have...

n = PV / RT = (1 atm)(63 L) / (0.0821 Latm/Kmol)(1273K)

n = 0.603 moles = 0.60 moles of CO₂ (2 sig. figs.)

Next, we use the stoichiometry of the balanced equation to find moles of acetylene (C₂H₂), and then convert moles to grams using the molar mass of C₂H₂:

0.60 mols CO₂ x 2 moles C₂H₂ / 4 mols CO₂ = 0.30 moles C₂H₂ needed

0.30 mols C₂H₂ x 26 g C₂H₂ / mole = 7.8 g C₂H₂ needed (2 sig. figs.)

To find the grams of water produced, we will use the stoichiometry of the balanced equation. Since we now know that 0.60 moles of CO₂ are produced, we use this value to find moles of H₂O, and then convert that to grams:

0.60 moles CO₂ x 2 mols H₂O / 4 moles CO₂ = 0.302 moles H₂O

0.302 moles H₂O x 18 g H₂O / mole = 5.4 g H₂O produced (2 sig. figs.)