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Kipish [7]
1 year ago
8

How many moles of H2O are produced from 6.23g of C2H6 in the following reaction? 2C2H6+7O2 -> 4CO2+6H2O

Chemistry
2 answers:
sweet-ann [11.9K]1 year ago
8 0

C2H6 (g) + O2 (g). --> CO2(g)+H2O (g)

2C2H6 (g) + 7O2 (g). --> 4CO2(g)+6H2O (g)

(8.45 x 10^23)/(6.022x10^23) = 1.403 moles H2O

1.403 moles H2O x (4molCO2/6molH2O)= .935 moles CO2

.935 moles CO2 x (6.022 x10^23)= 5.63 x 10^23 molecules CO2

azamat1 year ago
8 0

Answer:

0.623 moles of H₂O.

Explanation:

  • balanced equation: 2C₂H₆ + 7O₂ -> 4CO₂+ 6H₂O

Given:

  • 6.23g of C₂H₆
  • C₂H₆ molar mass is 30

solve for moles of C₂H₆

  • moles = mass/mr
  • moles = 6.23/30
  • moles = 0.2077

solve for moles of H₂O using molar ratio

  • 2C₂H₆ : 6H₂O
  • 2 : 6
  • 0.2077 : 0.623

Therefore, found that 0.623 moles of H₂O is produced.

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30. a. How do the properties of metals differ from those
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Explanation:

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3 years ago
The specific heat of aluminum is 0.897 J/g•°C. Which equation would you use to calculate the amount of heat needed to raise the
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6 0
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In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691
andre [41]

Explanation:

Let us assume that the value of K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s

Also at 1500 K, K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s

                     K_{r} = 400.613 m^{6}/mol^{2}s

Relation between K_{p} and K_{c} is as follows.

                  K_{p} = K_{c}RT

Putting the given values into the above formula as follows.

                  K_{p} = K_{c}RT

         0.003691 = K_{c} \times 8.314 \times 1500

                K_{c} = 2.9 \times 10^{-7}

Also,     K_{c} = \frac{K_{f}}{K_{r}}

or,                K_{f} = K_{c} \times K_{r}

                               = 2.9 \times 10^{-7} \times 400.613

                               = 1.1617 \times 10^{-4} m^{6}/mol^{2}s

Thus, we can conclude that the value of K_{f} is 1.1617 \times 10^{-4} m^{6}/mol^{2}s.

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