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Maslowich
2 years ago
8

In the presence of excess oxygen, methane gas burns in a constant-pressure system to yield carbon dioxide and water: CH4 (g) + 2

O2 (g) → CO2 (g) + 2H2O (l) △H = -890.0 kJ Calculate the value of q (kJ) in this exothermic reaction when 1.80 g of methane is combusted at constant pressure.
Chemistry
1 answer:
djyliett [7]2 years ago
3 0

Answer:

-100.125

Explanation:

We are given

          CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)    △H = -890.0 kJ/mol

The given information is for complete reaction

we have 1.8grams of Methane

Molar mass of CH4 = 16  

No of moles of Methane = 1.8/ 16

                                          = 0.1125

So the amount of the heat will be released from the amount of the methane

Q = No of moles*( -890.0 kJ/mol)

  = 0.1125*(-890)

   = -100.125kJ/mole

Therefore the amount of Energy released from 1.8 grams of methane is equal to -100.125

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How many liters of ammonia (NH3), at 3.2 atm and 23C, must be used to produce of 2.65 grams of calcium hydride (CaH2). 6 Ca(s)
7nadin3 [17]

Answer:

The answer to your question is    V = 0.32 L

Explanation:

Data

Volume of NH₃ = ?

P = 3.2 atm

T = 23°C

mass of CaH₂ = 2.65 g

Balanced chemical reaction

               6Ca  +  2NH₃   ⇒   3CaH₂  +  Ca₃N₂

Process

1.- Convert the mass of CaH₂ to moles

-Calculate the molar mass of CaH₂

 CaH₂ = 40 + 2 = 42 g

                             42 g ------------------ 1 mol

                              2.65 g --------------  x

                              x = (2.65 x 1)/42

                              x = 0.063 moles

2.- Calculate the moles of NH₃

                     2 moles of NH₃ --------------- 3 moles of CaH₂

                      x                        --------------- 0.063 moles

                                x = (0.063 x 2) / 3

                                x = 0.042 moles of NH₃

3.- Convert the °C to °K

Temperature = 23°C + 273

                      = 296°K

4.- Calculate the volume of NH₃

-Use the ideal gas law

              PV = nRT

-Solve for V

                V = nRT / P

-Substitution

                V = (0.042)(0.082)(296) / 3.2

-Simplification

               V = 1.019 / 3.2

-Result

               V = 0.32 L

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8 0
2 years ago
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