Question

q29. (7 points)
(a) add the coefficients to balance the following chemical equation (doing a tally will be helpful):
__ c₃h₈(g) + o₂(g) → co₂(g) + __ h₂o(l)
(b) how many moles of carbon dioxide are produced for every 24.3 moles of propane fuel (c₃h₈) burned? ____ mol
(c) what is the mass of carbon dioxide formed? show working, use part (b).
mm(co₂) = ____ g/mol
mass, m =

Explanation:

(a)

Step1: Balance carbon atoms

There are 3 carbon atoms in \(C_3H_8\). So, put a coefficient of 3 in front of \(CO_2\).
\(C_3H_8(g)+O_2(g)\to3CO_2(g)+H_2O(l)\)

Step2: Balance hydrogen atoms

There are 8 hydrogen atoms in \(C_3H_8\). So, put a coefficient of 4 in front of \(H_2O\).
\(C_3H_8(g)+O_2(g)\to3CO_2(g)+4H_2O(l)\)

Step3: Balance oxygen atoms

On the right - hand side, there are \(3\times2 + 4\times1=6 + 4 = 10\) oxygen atoms. So, put a coefficient of 5 in front of \(O_2\).
\(C_3H_8(g)+5O_2(g)\to3CO_2(g)+4H_2O(l)\)

Step1: Set up the proportion

Let \(n(CO_2)\) be the moles of \(CO_2\). Using the mole ratio \(\frac{n(CO_2)}{n(C_3H_8)}=\frac{3}{1}\)

Step2: Substitute the value of \(n(C_3H_8)\)

Given \(n(C_3H_8) = 24.3\) mol. Then \(n(CO_2)=3\times n(C_3H_8)\)

\(n(CO_2)=3\times24.3\) mol

Step1: Calculate the molar mass of \(CO_2\)

The molar mass of \(C\) is \(12.01\) g/mol and the molar mass of \(O\) is \(16.00\) g/mol.
\(MM(CO_2)=12.01+2\times16.00=12.01 + 32.00=44.01\) g/mol

Step2: Use the formula \(m = n\times MM\)

We know \(n(CO_2)=72.9\) mol (from part (b)) and \(MM(CO_2) = 44.01\) g/mol.
\(m(CO_2)=n(CO_2)\times MM(CO_2)\)
\(m(CO_2)=72.9\times44.01\)
\(m(CO_2)=72.9\times(44 + 0.01)=72.9\times44+72.9\times0.01\)
\(m(CO_2)=3207.6+0.729\)
\(m(CO_2)=3208.329\approx3208\) g

Answer:

\(1C_3H_8(g)+5O_2(g)\to3CO_2(g)+4H_2O(l)\)

(b)

From the balanced equation \(C_3H_8(g)+5O_2(g)\to3CO_2(g)+4H_2O(l)\), the mole ratio of \(C_3H_8\) to \(CO_2\) is \(1:3\).