Answer to Question #127103 in General Chemistry for azman

Question

Question 3

Answer the following questions by referring to the reaction stoichiometry given;
C3H8 + 5 O2 3 CO2 + 4 H2O
a)Calculate the mass of H2O is produced from the reaction of 6.3 g of propane.
b)Calculate how many molecules of H2O are produced when 2 moles of O2 are reacted with an excess of propane.
c)Calculate how many atoms of hydrogen in water are produced when 2.4 moles of propane are reacted.
d)Calculate how many molecules of CO2 are produced when 2.3 x 106 atoms of O2 are reacted.
e)Calculate the mass of CO2 produced when 6.5 g of propane is reacted with 14.2 g of O2.

Answer 1

a)

$$n(H_2O) = 4n(C_3H_8) = 4(m(C_3H_8)/M(C_3H_8))$$

$$n(H_2O) = 4*(6.3/44.1) = 0.57 mol$$

$$m(H_2O) = nM = 0.57*18.01 = 10.3 g$$

b)

$$n(H_2O) = 1.25n(O_2) = 1.25*2 = 2.5 mol$$

$$N(H_2O) = nN_A = 2.5*6.02*10^{23} = 15.05*10^{23}$$

c)

$$n(H_2O) = 4n(C_3H_8) = 4*2.4 = 4.8 mol$$

$$N(H_2O) = nN_A = 4.8*6.02*10^{23} = 28.896*10^{23}$$

$$N(H) = 2N(H_2O) = 2*28.896*10^{23} = 57.792*10^{23}$$

d)

$$n(O_2) = N/N_A = 2.3*10^6/6.02*10^{23} = 3.82*10^{-18}mol$$

$$n(CO_2) = 5/3n(O_2) = 6.37*10^{-18} mol$$

$$N(CO_2) = nN_A = 6.37*10^{-18}*6.02*10^{23} = 38.35*10^5 = 3.835*10^6$$

e)

$$n(C_3H_8) = m/M = 6.5/44.1 = 0.15 mol$$

$$n(O_2) = m/M = 14.2/32.0 = 0.44375 mol$$

O₂ in excess.

$$n(CO_2) = 3n(C_3H_8) = 3*0.15 = 0.45 mol$$

$$m(CO_2) = nM = 0.45*44.01 = 19.845 g$$