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saveliy_v [14]
3 years ago
14

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is

mixed with 12. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Round your answer to significant digits
2 C2H6 + 7 O2 = 4 CO2 + 6 H2O
Chemistry
1 answer:
Pie3 years ago
7 0

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

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Answer:

a) 2-bromopyrrole

Explanation:

Our options for this questions are:

a) 2-bromopyrrole

b) 2,3-dibromopyrrole

c) N-bromopyrrole

d) 3-bromopyrrole

To understand how the reaction works we have to start with the <u>resonance structures</u>. (Figure 1), on these structures, we will obtain a n<u>egative charge on carbon 2</u> in the pyrrole ring, therefore on this carbon we can generate an attack to an electrophile.

The second step is to check how the mechanism take place. An <u>electrophile is generated</u> by the Br_2 and FeBr_3. This electrophile can be <u>attacked</u> by the negative charge on carbon 2 producing the 2-bromopyrrole. (See figure 2).

I hope it helps!

5 0
3 years ago
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Murrr4er [49]

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lozanna [386]

The number  of calories that are  required  to change the temperature  of 2.18 g of water from 15.3 c to 69.5 c is  <u>118.16 cal</u>


    <u><em> calculation</em></u>

  •    Heat in calories  = MCΔ T where,
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3 years ago
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