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miv72 [106K]
3 years ago
9

How many moles are present in

Chemistry
1 answer:
olasank [31]3 years ago
4 0

Answer: (a) There are 0.428 moles present in 12 g of N_{2} molecule.

(b) There are 2 moles present in 12.044 \times 10^{23} particles of oxygen.

Explanation:

(a). The mass of nitrogen molecule is given as 12 g.

As the molar mass of N_{2} is 28 g/mol so its number of moles are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{12 g}{28 g/mol}\\= 0.428 mol

So, there are 0.428 moles present in 12 g of N_{2} molecule.

(b). According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms.

Therefore, moles present in 12.044 \times 10^{23} particles are calculated as follows.

Moles = \frac{12.044 \times 10^{23}}{6.022 \times 10^{23}}\\= 2 mol

So, there are 2 moles present in 12.044 \times 10^{23} particles of oxygen.

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The answer to your question is 98.9 %

Explanation:

Data

moles of methane = CH₄ = 2.0

excess air

Percent yield = ?

mass of CO₂ = 87 g

- Balanced chemical reaction

                CH₄  +  2O₂   ⇒  CO₂  +  2H₂O

      Reactants     Elements       Products

             1                    C                   1

             4                   H                   2

             4                   O                   2

- Calculate the molar mass of CH₄

CH₄ = 12 + 4 = 16 g

- Convert the moles to mass

                     16 g of CH₄ ----------------- 1 mol

                       x                -----------------  2 moles

                       x = (2 x 16) / 1

                       x = 32 g of CH₄

-Calculate the theoretical formation of CO₂

                     16 g of CH₄ ----------------- 44 g of CO₂

                     32 g of CH₄ ----------------  x

                            x = (32 x 44) / 16

                            x = 88 g of CO₂

-Calculate the Percent yield

     Percent yield = Actual yield/Theoretical yield x 100

     Percent yield = 87/88 x 100

    Percent yield = 98.9 %

3 0
3 years ago
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