Question

2 C₂H₂ (g) + 5 O₂ (g) à 4 CO₂ (g) + 2 H₂O (g)

How many moles of O2 are needed to react with 15.0 mL of C2H2 at 125oC and 1.25 atm? Please show work!

Answer 1

PV = nRT

where, P = pressure = 1.25 atm

V = volume = 15.0 mL = 0.015 L

n = number of moles

R = Gas constant

T = temperature = 125 + 273 = 398 K

1.25 * 0.015 = n * 0.0821 * 398

0.0188 = n * 32.7

n = 0.0188 / 32.7 = 5.75*10^-4 mole

Number of moles of C2H2 = 5.75*10^-4 mole

From the balanced equation we can say that

2 mole of C2H2 requires 5 mole of O2 so

5.75*10^-4 mole of C2H2 will require

= 5.75*10^-4 mole of C2H2 *(5 mole of O2 / 2 mole of C2H2)

= 14.4*10^-4 mole of O2

Therefore, the number of moles of O2 required would be 0.00144 mole