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Gala2k [10]
3 years ago
10

Write the stracture of 2,2-dimetyl butane​

Chemistry
1 answer:
Mice21 [21]3 years ago
6 0

Answer:

Explanation:

Let's break it down:

  • Butane- A 4 carbon alkane
  • Methyl - A CH₃ group
  • 2,2 - Both the Methyl groups are attached at carbon 2

We can now draw the structure of  2,2-dimethyl butane​ using the following steps:

  1. Draw the backbone (butane)
  2. Draw the substituents (2 methyl groups at carbon 2)

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The reaction of NH3 and O2 forms NO and water. The NO can be used to convert P4 to P4O6, forming N2 in the process. The P4O6 can
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Answer : The mass of PH_3 produced from the reaction is, 0.651 grams.

Explanation :

The chemical reactions used are:

(1) 4NH_3+5O_2\rightarrow 4NO+6H_2O

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(4) 4H_3PO_4\rightarrow PH_3+3H_3PO_4

First we have to calculate the moles of NH_3

\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}

Molar mass of NH_3 = 17 g/mol

\text{Moles of }NH_3=\frac{1.95g}{17g/mol}=0.115mol

Now we have to calculate the moles of NO

From the balanced chemical reaction 1, we conclude that:

As, 4 moles of NH_3 react to give 4 moles of NO

So, 0.115 moles of NH_3 react to give 0.115 moles of NO

Now we have to calculate the moles of P_4O_6

From the balanced chemical reaction 2, we conclude that:

As, 6 moles of NO react to give 1 moles of P_4O_6

So, 0.115 moles of NO react to give \frac{0.115}{6}=0.0192 moles of P_4O_6

Now we have to calculate the moles of H_3PO_4

From the balanced chemical reaction 3, we conclude that:

As, 1 moles of P_4O_6 react to give 4 moles of H_3PO_4

So, 0.0192 moles of P_4O_6 react to give 0.0192\times 4=0.0768 moles of H_3PO_4

Now we have to calculate the moles of PH_3

From the balanced chemical reaction 4, we conclude that:

As, 4 moles of H_3PO_4 react to give 1 moles of PH_3

So, 0.0768 moles of H_3PO_4 react to give \frac{0.0768}{4}=0.0192 moles of PH_3

Now we have to calculate the mass of PH_3

\text{ Mass of }PH_3=\text{ Moles of }PH_3\times \text{ Molar mass of }PH_3

Molar mass of PH_3 = 33.9 g/mole

\text{ Mass of }PH_3=(0.0192moles)\times (33.9g/mole)=0.651g

Therefore, the mass of PH_3 produced from the reaction is, 0.651 grams.

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