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seropon [69]
3 years ago
5

Write the coefficients for the reactant and product.

Chemistry
2 answers:
Sergeu [11.5K]3 years ago
8 0

Answer:

2, 7, 4, 6

2C₂H₆  + 7O₂ → 4CO₂ + 6H₂O

Explanation:

The given reaction is combustion of ethane. Ethane burns in the presence of oxygen and gives carbondioxide and water.

Chemical equation:

C₂H₆  + O₂ → CO₂ + H₂O

Balanced Chemical equation:

2C₂H₆  + 7O₂ → 4CO₂ + 6H₂O

C₂H₆  + O₂                          →                CO₂ + H₂O

In reactant side                                    in product side

C = 2                                                          C = 1

H = 6                                                          H = 2

O = 2                                                          O = 3

Multiply the reactant side C₂H₆ with 4 and O₂ 7.

2C₂H₆  + 7O₂                          →                CO₂ + H₂O

C = 4

H = 12

O = 14

Now multiply the CO₂ and  H₂O with a coefficient so that C =4, H = 12 and O = 14

2C₂H₆  + 7O₂                          →                4CO₂ + 6H₂O

C = 4                                                            C= 4

H = 12                                                         H= 12

O = 14                                                         O=14

balu736 [363]3 years ago
4 0
<h3>Answer:</h3>

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

<h3>Explanation:</h3>
  • The equation for the reaction shown represents the combustion of ethane.

C₂H₆ + O₂ → CO₂ + H₂O

  • To balance the equation we put the coefficients, 2, 7, 4, and 6 on the reactants and products.

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

  • Balancing chemical equations makes the number of atoms of each element equal on both side of the equation.
  • It ensures that chemical equations obey the law of conservation of mass in chemical reactions.
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Radda [10]

Answer: (1). There are  0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP.

(2). The density of gaseous arsine is 3.45 g/L.

Explanation:

1). At STP the pressure is 1 atm and temperature is 273.15 K. So, using the ideal gas equation number of moles are calculated as follows.

PV = nRT

where,

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Substitute the values into above formula as follows.

PV = nRT\\1 atm \times 0.372 L = n \times 0.0821 L atm/mol K \times 273.15 K\\n = 0.0165 mol

2). As number of moles are also equal to mass of a substance divided by its molar mass.

So, number of moles of Arsine (AsH_{3}) (molar mass = 77.95 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\0.0165 mol = \frac{mass}{77.95 g/mol}\\mass = 1.286 g

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Density = \frac{mass}{volume}\\= \frac{1.286 g}{0.372 L}\\= 3.45 g/L

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Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

H2SO3\leftrightarrow H^{+}+HSO3^{-} --------(1)

where,  Ka1 = 1.5 x 10–2  or pKa1 = 1.824

HSO3^{-}\leftrightarrow H^{+}+SO3^{2-} --------(2)

where,  Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

i.e. \ 5.13\  moles/L*0.385L = moles\  NaOH\\therefore, \ moles\  NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L

Step 2:

For the second equivalence point setup an ICE table:

                  HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}

Initial           1.98                    ?                                       0

Change      -x                       -x                                       x

Equil           1.98-x                 ?-x                                    x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles

Volume of NaOH required is:

\frac{0.634\ moles}{0.615 moles/L}=0.389L

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

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