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MaRussiya [10]
3 years ago
10

1. How many moles of oxygen gas are required to react with 15.56 moles of hydrogen gas? (2 pts.)

Chemistry
1 answer:
loris [4]3 years ago
7 0

Answer:

7.78moles of O2

Explanation:

2H2 (g) + O2 (g) → 2H2O (g)

From the balanced equation,

2 moles of H2 required 1mole of O2.

Therefore, 15.56 moles of H2 will require = 15.56/2 = 7.78moles of O2

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Which of the following best describes the distribution of the world’s food supply? Select all that apply.
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Explanation: I did the test and got it right. Hope this helps!

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Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the t
Sladkaya [172]

Explanation:

The given balanced reaction is as follows.

           2NH_{4}NO_{3} \rightarrow 2N_{2} + O_{2} + 4H_{2}O

It is given that mass of ammonium nitrate is 86.0 kg.

As 1 kg = 1000 g. So, 86.0 kg = 86000 g.

Hence, moles of NH_{4}NO_{3} present will be as follows.

      Moles of NH_{4}NO_{3} = \frac{mass given}{\text{molar mass of NH_{4}NO_{3}}}

                                  = \frac{86000 g}{80.043 g/mol}

                                  = 1074.42 mol

Therefore, moles of N_{2}, O_{2} and H_{2}O produced by 1074.42 mole of NH_{4}NO_{3} will be as follows.

  Moles of O_{2} = \frac{1}{2} \times 1074.42 mol

                                = 537.21 mol

Moles of N_{2} = \frac{2}{2} \times 1074.42 mol

                                = 1074.42 mol

Moles of H_{2}O = \frac{4}{2} \times 1074.42 mol

                                = 2148.84 mol

Therefore, total number of moles will be as follows.

          537.21 mol + 1074.42 mol + 2148.84 mol

        = 3760.47 mol

According to ideal gas equation, PV = nRT. Hence, calculate the volume as follows.

                       PV = nRT

                     1 atm \times V = 3760.47 mol \times 0.0821 L atm/mol K \times 580 K[/tex]         (as 307^{o}C = 307 + 273 = 580 K)

                           V = 179066.06 L

Thus, we can conclude that total volume of the gas is 179066.06 L.

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