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fenix001 [56]
3 years ago
15

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

nitrogen, hydrogen is considered the limiting reactant because 7. 5 mol of hydrogen would be needed to consume the available nitrogen. 7. 5 mol of nitrogen would be needed to consume the available hydrogen. Hydrogen would produce 7. 5 mol more ammonia than nitrogen. Nitrogen would produce 7. 5 mol more ammonia than hydrogen.
Chemistry
1 answer:
harkovskaia [24]3 years ago
6 0

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g)   (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

n = \frac{m}{M}

Where:    

m: is the mass

M: is the molar mass

  • For <em>hydrogen </em>we have:

n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles

  • And for <em>nitrogen</em>:

n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles

We can see in reaction (1) that <u>3 moles of hydrogen</u> react with <u>1 mol of nitrogen</u>, so the number of hydrogen moles needed to react nitrogen is:

n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles

Since we have <u>3.47 moles of hydrogen</u> and we need <u>7.50 moles</u> to react with all the mass of nitrogen, the <em>limiting reactant</em> is <em>hydrogen</em>.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that <u>3 moles of hydrogen</u> produces <u>2 moles of ammonia</u>, so:

n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

Therefore, hydrogen is the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

   

I hope it helps you!                        

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Answer:

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Explanation:

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Potassium oxide: K₂O.

There's no need for prefixes since K₂O is an ionic compound.

<h3>Explanation</h3>

Find the two elements on a periodic table:

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  • Oxygen- O- near the right end of periodic two.

Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.

  • Potassium is a metal,
  • Oxygen is a nonmetal.

A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:

  • Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.
  • Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion \text{O}^{2-} when it combines with metals.

The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each \text{O}^{2-} ion is twice that on a \text{K}^{+} ion. Each \text{K}^{+} would pair up with two \text{O}^{2-}. Hence the subscript in the formula: \text{K}_{\bf 2}\text{O}.

There are two classes of compounds:

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Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide \text{CO}_2, the prefix di- indicates that there are two oxygen atoms in the formula \text{CO}_2. However, there's no need for prefix in ionic compounds such as \text{K}_2\text{O}.

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