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Answer:
[tex]\begin{gathered} 1)3moles_{} \\ 2)\text{ }592.38grams \end{gathered}[/tex]Explanations:
a) To get the number of moles of CO2 that are produced for each mole of C3H8 used, we will use the stochiometry ratio from the balanced chemical reaction.
The balanced chemical reaction that resulted from the combustion of propane gas is given as:
[tex]C_3H_8+5O2\to3CO_2+4H_2O[/tex]From the balanced reaction, you can see based on stoichiometry that for each mole of propane that reacted, 3 moles of CO2 was produced.
b) If you start with 362 grams of C3H8, the number of moles of propane will be expressed according to the formula:
[tex]\text{moles of C}_3H_8=\frac{Mass}{Mola\text{r mass}}\text{ }[/tex]Given the following parameters:
Mass of propane gas = 362 grams
Molar mass of propane = (3*12) + (1*8) = 36 + 8 = 44g/mol
[tex]\begin{gathered} \text{Moles of C}_3H_8=\frac{362}{44} \\ \text{Moles of C}_3H_8=8.227moles \end{gathered}[/tex]From stoichiometry, you can see that 1 mole of C3H8 produces 4 moles of H2O, hence the moles of H2O that will be produced is given as:
[tex]\begin{gathered} \text{Moles of H}_2O=4\times8.227 \\ \text{Moles of H}_2O\approx32.91\text{ moles} \end{gathered}[/tex]Get the mass of H2O that was produced using the formula:
[tex]\begin{gathered} \text{Mass of H}_2O=moles\text{ }\times molar\text{ mass} \\ \text{Mass of H}_2O=32.91\cancel{\text{moles}}\times\frac{18g}{\cancel{\text{mol}}} \\ \text{Mass of H}_2O=32.91\times18g \\ \text{Mass of H}_2O=592.38\text{grams} \end{gathered}[/tex]Therefore if you start with 362 grams of C3H8, 592.38grams of H2O will be produced
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