Limiting reagent problem? For the reaction $C_{2} H_{6} + O_{2} \to C O_{2} + H_{2} O$ $C_2H_6+O_2 to CO_2+H_2O$

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Limiting reagent problem? For the reaction $C_{2} H_{6} + O_{2} \to C O_{2} + H_{2} O$ $C_2H_6+O_2 to CO_2+H_2O$

(1) Balance
(2) if $270 g$ $"270 g"$ $C_{2} H_{6}$ $"C"_2"H"_6$ and $300 g$ $"300 g"$ $O_{2}$ $"O"_2$ reacted, how many moles of ${CO}_{2}$ $"CO"_2$ were produced?
(3) How many grams excess leftover?

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Answer 1 · 2016-02-16T05:04:21

see below

Explanation:

Balanced Eqn
$2 C_{2} H_{6} + 7 O_{2} = 4 C O_{2} + 6 H_{2} O$ $2C_2H_6 +7O_2 =4 CO_2+6H_2O$
By the Balanced eqn
60g ethane requires 7x32= 224g oxygen
here ethane is in excess.oxygen will be fully consumed
hence
300g oxygen will consume $60 \cdot \frac{300}{224} = 80.36 g$ $60*300/224 = 80.36 g$ ethane
leaving (270-80.36)= 189.64 g ethane.
By the Balanced eqn
60g ethane produces 4x44 g CO2
hence amount of CO2 produced = $4 \cdot 44 \cdot \frac{80.36}{60} = 235.72 g$ $4*44*80.36/60 =235.72g$

and its no. of moles will be $\frac{235.72}{44}$ $235.72/44$ =5.36 where 44 is the molar mass of Carbon dioxide

Answer 2 · 2016-02-18T03:34:21

Balanced equation is $C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O$ $C_2H_6+7/2O_2 to 2CO_2+3H_2O$

Explanation:

Balanced equation is $C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O$ $C_2H_6+7/2O_2 to 2CO_2+3H_2O$

Alternatively, this can be written as $2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O$ $2C_2H_6+7O_2 to 4CO_2+6H_2O$

Find the number of moles of each reactant:

For $C_{2} H_{6}$ $C_2H_6$ , with molar mass $30$ $30$ $g m o l^{- 1}$ $gmol^-1$ :

$n = \frac{m}{M} = \frac{270}{30} = 9$ $n=m/M=270/30=9$ $m o l$ $mol$

For $O_{2}$ $O_2$ , with molar mass $32$ $32$ $g m o l^{- 1}$ $gmol^-1$ :

$n = \frac{m}{M} = \frac{300}{32} = 9.375$ $n=m/M=300/32=9.375$ $m o l$ $mol$

It might look as though we have an excess of $O_{2}$ $O_2$ , but remember that each $1$ $1$ $m o l$ $mol$ of $C_{2} H_{6}$ $C_2H_6$ required $\frac{7}{2}$ $7/2$ $m o l$ $mol$ of $O_{2}$ $O_2$ , so in fact we have less oxygen than we need. Some $C_{2} H_{6}$ $C_2H_6$ will remain unburned at the end of the reaction.

Divide the $9.375$ $9.375$ $m o l$ $mol$ by $\frac{7}{2}$ $7/2$ to discover that $2.68$ $2.68$ $m o l$ $mol$ of $C_{2} H_{6}$ $C_2H_6$ will react. Each mole of $C_{2} H_{6}$ $C_2H_6$ produces $2$ $2$ $m o l$ $mol$ of $C O_{2}$ $CO_2$ , so $5.36$ $5.36$ $m o l$ $mol$ of $C O_{2}$ $CO_2$ will be produced.

The remaining amount of $C_{2} H_{6}$ $C_2H_6$ will be $9 - 2.68 = 6.32$ $9-2.68=6.32$ $m o l$ $mol$ . To find the mass of excess $C_{2} H_{6}$ $C_2H_6$ :

$m = n M = 6.32 \cdot 30 = 189.6$ $m=nM=6.32*30=189.6$ $g$ $g$

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Limiting reagent problem? For the reaction $C_{2} H_{6} + O_{2} \to C O_{2} + H_{2} O$ $C_2H_6+O_2 to CO_2+H_2O$

(1) Balance
(2) if $270 g$ $"270 g"$ $C_{2} H_{6}$ $"C"_2"H"_6$ and $300 g$ $"300 g"$ $O_{2}$ $"O"_2$ reacted, how many moles of ${CO}_{2}$ $"CO"_2$ were produced?
(3) How many grams excess leftover?

2 Answers

Explanation:

Explanation:

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Limiting reagent problem? For the reaction C2H6+O2→CO2+H2OC_2H_6+O_2 to CO_2+H_2O

(1) Balance (2) if 270 g"270 g" C2H6"C"_2"H"_6 and 300 g"300 g" O2"O"_2 reacted, how many moles of CO2"CO"_2 were produced? (3) How many grams excess leftover?

2 Answers

Explanation:

Explanation:

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Limiting reagent problem? For the reaction $C_{2} H_{6} + O_{2} \to C O_{2} + H_{2} O$ $C_2H_6+O_2 to CO_2+H_2O$

(1) Balance
(2) if $270 g$ $"270 g"$ $C_{2} H_{6}$ $"C"_2"H"_6$ and $300 g$ $"300 g"$ $O_{2}$ $"O"_2$ reacted, how many moles of ${CO}_{2}$ $"CO"_2$ were produced?
(3) How many grams excess leftover?