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Arturiano [62]
3 years ago
8

Hydrogenation reactions, in which H2 and an "unsaturated" organic compound combine, are used in the food, fuel, and polymer indu

stries. In the simplest case, ethene (C2H4) and H2 form ethane (C2H6). If 140 kJ is given off per mole of C2H4 reacting, How much heat (in MJ) is released when 12 kg of C2H6 forms?
Chemistry
1 answer:
ddd [48]3 years ago
8 0

<u>Answer:</u> The amount of heat released is 56 MJ.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

Given mass of C_2H_6 = 12 kg = 12000 g    (Conversion factor: 1 kg = 1000 g)

Molar mass of C_2H_6 = 30 g/mol

Putting values in above equation, we get:

\text{Moles of }C_2H_6=\frac{12000g}{30g/mol}=400mol

The chemical reaction for hydrogenation of ethene follows the equation:

C_2H_4+H_2\rightarrow C_2H_6

By Stoichiometry of the reaction:

When 1 mole of ethane releases 140 kJ of heat.

So, 400 moles of ethane will release = \frac{140}{1}\times 400=56000kJ of heat.

Converting this into Mega joules, using the conversion factor:

1 MJ = 1000 kJ

So, \Rightarrow 56000kJ\times (\frac{1MJ}{1000kJ})=56MJ

Hence, the amount of heat released is 56 MJ.

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Read 2 more answers
How many liters of hydrogen gas will be produced at STP from the reaction of 7.179 x 1023 atoms of magnesium with 54.21 g of pho
wlad13 [49]
<h3>Answer:</h3>

18.58 liters of hydrogen gas

<h3>Explanation:</h3>

We are given;

  • The equation;

3Mg + 2H₃(PO₄) → Mg₃(PO₄)₂ + 3H₂

  • Atoms of Magnesium = 7.179 x 10^23 atoms
  • Mass of phosphoric acid as 54.21 g

We are required to determine the volume of hydrogen gas produced;

Step 1; moles of Magnesium

1 mole of an element contains 6.02 × 10^23 atoms

therefore;

Moles of Mg = (7.179 x 10^23 ) ÷ (6.02 × 10^23)

                   = 1.193 moles

Step 2: Moles of phosphoric acid

moles = Mass ÷ Molar mass

Molar mass of phosphoric acid = 97.994 g/mol

Therefore;

Moles of Phosphoric acid = 54.21 g ÷ 97.994 g/mol

                                           = 0.553 moles

Step 3: Determine the rate limiting reagent

From the mole ratio of Mg to Phosphoric acid (3 : 2);

1.193 moles of magnesium requires 0.795 moles of phosphoric acid while,

0.0553 moles of phosphoric acid requires 0.8295 moles of Mg

Therefore, phosphoric acid is the rate limiting reagent

step 4: Determine the moles of hydrogen produced

From the equation, w moles of phosphoric acid reacts to produce 3 moles of hydrogen;

Therefore; moles of Hydrogen = moles of phosphoric acid × 3/2

                                                   = 0.553 moles × 3/2

                                                   = 0.8295 moles

Step 5: Volume of hydrogen gas

1 mole of a gas occupies a volume of 22.4 liters at STP

Therefore;

Volume of Hydrogen = 0.8295 moles × 22.4 L/mol

                                  = 18.58 Liters

Therefore; 18.58 liters of hydrogen gas  will be produced

4 0
3 years ago
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