
Using the balanced equation shown below, what is the mass of $ {C_3}{H_8} $ that must react in order to release $ 1.25 \times {10^6}{\text{kJ}} $ of heat? $ \Delta {H_{rxn}} = - 2219.9{\text{kJ}} $
Answer
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Hint: We are already provided with the compound. We know the mass of $ {C_3}{H_8} $ . Also, we know the amount of heat released. From this value using the formula we can find the number of moles to further find mass.
Formula used: $ {\text{moles = }}\dfrac{{{\text{ given mass }}}}{{{\text{molar mass}}}} $
Complete step by step solution:
We already know that Reactions tend to occur at constant pressure, so consequently, we write that
$ {Q_{rxn}} = \Delta {H_{rxn}} $
Also,
$ {\overline {\Delta H} _{rxn}} = \dfrac{{\Delta {H_{rxn}}}}{{{n_{{C_3}{H_8}}}}} $
So, the moles of the reaction from the above equations can be written as:
$ {n_{{C_3}{H_8}}} = - 1.25 \times {10^6}{\text{kJ}} \times \dfrac{{1{\text{mol}}}}{{ - 2219.9{\text{kJ}}}} $
$ {n_{{C_3}{H_8}}} = 563.09{\text{mol}} $
Now we know the moles of compound, it is simple to find the mass:
$ {\text{moles = }}\dfrac{{{\text{ given mass }}}}{{{\text{molar mass}}}} $
On putting the value in the equation,
$ {\text{mass}} = 563.09{\text{mol}} \times 44.1{\text{g/mol}} $
$ {\text{mas}}{{\text{s}}_{{C_3}{H_8}}} = 24832.20{\text{g}} $
So, the weight of $ {C_3}{H_8} $ is $ 24.8{\text{ kg}} $ .
Additional Note: We should know that the compound used here is propane. It is made up only of carbon and hydrogen so it's an organic compound. We should know that in order to calculate the number of moles of any substance present in the sample, we simply divide the given weight of the substance by its molar mass.
One mole of any substance is equal to the value of $ 6.023 \times {10^{23}} $ (Avagadro number). It can be used to measure the products obtained from the chemical reaction. The unit is denoted by mol.
Note:
We already know that the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of the compound by the molar mass of the compound expressed in grams.
Some uses of propane are: its main uses include home and water heating, cooking and refrigerating food, clothes drying, and powering farm and industrial equipment. The chemical industry also uses propane as a raw material for making plastics and other compounds.
Formula used: $ {\text{moles = }}\dfrac{{{\text{ given mass }}}}{{{\text{molar mass}}}} $
Complete step by step solution:
We already know that Reactions tend to occur at constant pressure, so consequently, we write that
$ {Q_{rxn}} = \Delta {H_{rxn}} $
Also,
$ {\overline {\Delta H} _{rxn}} = \dfrac{{\Delta {H_{rxn}}}}{{{n_{{C_3}{H_8}}}}} $
So, the moles of the reaction from the above equations can be written as:
$ {n_{{C_3}{H_8}}} = - 1.25 \times {10^6}{\text{kJ}} \times \dfrac{{1{\text{mol}}}}{{ - 2219.9{\text{kJ}}}} $
$ {n_{{C_3}{H_8}}} = 563.09{\text{mol}} $
Now we know the moles of compound, it is simple to find the mass:
$ {\text{moles = }}\dfrac{{{\text{ given mass }}}}{{{\text{molar mass}}}} $
On putting the value in the equation,
$ {\text{mass}} = 563.09{\text{mol}} \times 44.1{\text{g/mol}} $
$ {\text{mas}}{{\text{s}}_{{C_3}{H_8}}} = 24832.20{\text{g}} $
So, the weight of $ {C_3}{H_8} $ is $ 24.8{\text{ kg}} $ .
Additional Note: We should know that the compound used here is propane. It is made up only of carbon and hydrogen so it's an organic compound. We should know that in order to calculate the number of moles of any substance present in the sample, we simply divide the given weight of the substance by its molar mass.
One mole of any substance is equal to the value of $ 6.023 \times {10^{23}} $ (Avagadro number). It can be used to measure the products obtained from the chemical reaction. The unit is denoted by mol.
Note:
We already know that the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of the compound by the molar mass of the compound expressed in grams.
Some uses of propane are: its main uses include home and water heating, cooking and refrigerating food, clothes drying, and powering farm and industrial equipment. The chemical industry also uses propane as a raw material for making plastics and other compounds.
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