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valentina_108 [34]
3 years ago
7

The combustion of ethane ( C 2 H 6 ) produces carbon dioxide and steam. 2 C 2 H 6 ( g ) + 7 O 2 ( g ) ⟶ 4 CO 2 ( g ) + 6 H 2 O (

g ) How many moles of CO 2 are produced when 5.20 mol of ethane is burned in an excess of oxygen?
Chemistry
2 answers:
GREYUIT [131]3 years ago
5 0

Answer:

10.4moles of CO2

Explanation:

2C2H6 + 7O2 —> 4CO2 + 6H2O

From the equation,

2moles of C2H6 produced 4 moles of CO2.

Therefore, 5.2moles of C2H6 will produce = (5.2 x 4)/2 = 10.4moles of CO2

dexar [7]3 years ago
4 0

Answer:

10.4 moles of CO2 are produced

Explanation:

take the 5.2 moles of C2H6 and multiply that by the mole ratio of CO2 to C2H6 in the reaction (4/2)

5.2 * (4/2) = 10.4

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Answer:

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If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?
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Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}  where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation:  T_C+273=T_K

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, T_1 = 273[K], and T_2 = (49.4+273)[K]=322.4[K].

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}

\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})

1810.04571428[mL]=V_2

Adjusting for significant figures, this gives V_2=1810[mL]

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