Question

Consider the balanced equation below C3H8O + 502 = 3CO2 + 4H2O A) How many grams of oxygen are needed to react with 26.4 mL of C3H8O (d= 786 Kg/m^3) B) How many grams of water and carbon dioxide will be formed if 26.4 mL of C3H80 are reacted with 1.20 moles of oxygen? How many grams of the reagent in excess are left? C) Assuming an 85% yeild, calculate the mass of water and the mass carbon dioxide formed

Answer 1

a,

mass of C3H8O = D*V = 26.4*0.786 = 20.7504 g

mol = mass/MW = 20.7504/60.095 = 0.34529

1 mol of C3H8O requires 5 mol, then 0.34529 mol --> 1.72645 mol of O2 required

mass = mol*MW = 1.72645*32 = 55.2464 g of O2

b)

if only 1.20 mol of O2, then O2 limits reaction

1.2 mol of O2 forms --> 3/5*mol of CO2 = 3/5*1.2 = 0.72 mol of H2O

mass = mol*MW = 0.72*44= 31.68 g of CO2

1.2 mol of O2 forms --> 4/5*mol of CO2 = 4/5*1.2 = 0.96mol of H2O

mass = mol*MW = 0.96*18= 17.28 g of water

c.

if 85% yield, then

mass = mol*MW = 0.72*44= 31.68 g of CO2 --> 85% --> 0.85*31.68 = 26.928 g of CO2

mass = mol*MW = 0.96*18= 17.28 g of water --> 85% --> 0.85*17.28 = 14.688 g of H2O