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3 years ago

Calculate the amount of CO2 produced when 1.00 mole of propane (C3H8) is burned in the presence of 1.00 mole of oxygen.

Chemistry

1 answer:

Anna [14]3 years ago

4 0

Answer:

The balanced Chemical equation is

$1C_3 H_8 +5O_2 >3CO_2+4H_2 O$

Since quantity of two reactants are given Hence we find the moles of product using both.

The quantity of product produced is the one we consider from the limiting reactant.

Limiting reactant is the reactant which runs out first.

So, least product is produced from the limiting reactant.

Mole ratio of $C_3 H_8:CO_2$ is 1: 3

$1.00 \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{\mathrm{8}} \times \frac{3 \mathrm{mol} \mathrm{CO}_{2}}{1 \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{8}}=3 \mathrm{molCO}_{2}$

Mole ratio of $O_2:CO_2$ is 5: 3

$1.00 \mathrm{mol} \mathrm{O}_{2} \times \frac{3 \mathrm{mol} \mathrm{co}_{2}}{5 \mathrm{mol} \mathrm{O}_{2}}=0.6 \mathrm{mol} \mathrm{CO}_{2}$

(Least produced)

So, the limiting reactant is $O_2$

The amount of $CO_2$ formed is 0.600mol (Answer)

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Read 2 more answers

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$