Question

Answer the following question 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 6.3g 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 x10 6 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

The balanced equation of the reaction is:

C₃H₈ +5 O₂ $\rightarrow$ 3 CO₂ + 4 H₂O

a) the mass of H₂O produced from the reaction of 6.3g of propane. According to the equation above, 1 mol of propane produces 4 mol of water:

$n(C_3H_8) = \frac{n(H_2O)}{4}$ .

The number of the moles of propane can be calculated using its mass (6.3 g) and its molar mass (44.1 g/mol):

$n(C_3H_8) = \frac{m}{M} = \frac{6.3}{44.1} = 0.143$ mol.

Therefore, the number of the moles of water produced is:

$n(H_2O) = 4\cdot0.143 = 0.571$ mol.

Finally, the mass of water is (M = 18.02 g/mol):

$m(H_2O) = 0.571\cdot 18.02 = 10.3$ g.

b) how many molecules of H₂O are produced when 2 moles of O₂ are reacted with an excess of propane. According to the reaction equation, when 5 mol of O₂ react, 4 mol of water is produced:

$\frac{n(O_2)}{5} = \frac{n(H_2O)}{4}$ .

Therefore, when 2 mol of O₂ reacts with an excess of propane, the number of the moles of water produced is:

$n(H_2O) = 4\cdot\frac{2}{5} = 1.6$ mol.

The number of the molecules in 1.6 mol can be calculated using the Avogadro number:

$N = n\cdot N_A = 1.6\cdot6.022\cdot10^{23} = 9.64\cdot10^{23}$ molecules.

c) how many atoms of hydrogen in water are produced when 2.4 moles of propane react.

Using the relation of part a):

$n(H_2O) = 4n(C_3H_8) = 4\cdot2.4 = 9.6$ mol.

In one molecule of water, there are 2 atoms of H. Thus, the number of hydrogen atoms is:

$N = 2N_A\cdot n(H_2O) = 2\cdot6.022\cdot10^{23}\cdot9.6$

$N = 1.16\cdot10^{25}$ atoms.

d) how many molecules of CO₂ are produced when 2.3x10⁶ atoms of O₂ are reacted. You mean 2.3x10⁶ atoms of O and not O₂, because O₂ is a molecule. Again,

$\frac{n(O_2)}{5} = \frac{n(CO_2)}{3}$ .

In one molecule of O₂ there are 2 oxygen atoms:

$n(CO_2) = 3\cdot\frac{2.3\cdot10^6}{2\cdot5\cdot N_A}$ mol.

Finally,

$N(CO_2) = n(CO_2)\cdot N_A = 3\cdot\frac{2.3\cdot10^6}{2\cdot5} = 6.9\cdot10^5$ molecules.

e) the mass of CO₂ produced when 6.5 g of propane is reacted with 14.2 g of O₂. Again,

$\frac{n(O_2)}{5} = \frac{n(C_3H_8)}{1}$ .

The number of the moles of O₂:

$n(O_2) = 14.2/31.998 = 0.444$ mol.

The number of the moles of propane:

$n(C_3H_8) = 6.5/44.1 = 0.147$ mol.

Which component is in excess? 0.444/5 = 0.089, 0.147/1 = 0.147, 0.089<0.147, so the propane is in excess. Therefore, we calculate the number of the moles and the mass of CO₂ produced using the mass and the number of the moles of O₂:

$\frac{n(O_2)}{5} = \frac{n(CO_2)}{3}$

$n(CO_2) = 3\cdot\frac{0.444}{5} = 0.266$ mol

$m(CO_2) = 0.266\cdot44.01 = 11.7$ g.

Answer: a) 10.3g, b) $9.64\cdot10^{23}$ molecules, c)  $1.16\cdot10^{25}$ atoms, d)$6.9\cdot10^5$ molecules, e)11.7 g.