Question

How many moles of carbon monoxide are required to react completely with 38.4 L of oxygen gas according to the following reaction at 0°C and 1 atm? carbon monoxide (g oygen(grbon dioxide (g) moles carbon monoxide How many moles of hydrogen peroxide (H202) are needed to produce 26.7 L of oxygen gas according to the following reaction at o °C and I atm? hydrogen peroxide (HiO2) (aq)water (+ oxgen(g) moles hydrogen peroxide (H202)

Answer 1

1)

The balanced equation is

2 CO + O2 -------> 2 CO2

PV = nRT

P = pressure = 1 atm

V = volume = 26.7 L

n = number of moles

R = Gas constant

T = temperature = 0 + 273 = 273 K

1 * 38.4 = n * 0.0821*273

38.4 = n * 22.4

n = 38.4 / 22.4 = 1.71 mole

From the balanced equation we can say that

1 mole of O2 requires 2 mole of CO so

1.71 mole of O2 will require

= 1.71 mole of O2 *(2 mole of CO / 1 mole of O2)

= 3.42 mole of CO

Therefore, the number of moles of CO = 3.42 mole

2) 2 H2O2 -----> 2 H2O + O2

PV = nRT

1 * 26.7 = n * 0.0821 * 273

26.7 = n * 22.4

n = 26.7 / 22.4 = 1.19 mol

number of moles of O2 = 1.19 mol

From the balanced equation we can say that

1 mole of O2 requires 2 mole of H2O2 so

1.19 mole of O2 will require

= 1.19 mole of O2 *(2 mole of H2O2 / 1 mole of O2)

= 2.38 mole of H2O2

mass of 1 mole of H2O2 = 34.0147 g

so the mass of 2.38 mole of H2O2 = 81.0 g

Therefore, the mass of H2O2 required would be 81.0 g