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NISA [10]
4 years ago
8

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t

he first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H 2 O H2O that can be produced by combining 62.8 g 62.8 g of each reactant? 4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Chemistry
1 answer:
ICE Princess25 [194]4 years ago
3 0

Answer:

42,3g of H₂O

Explanation:

For the reaction:

4NH₃(g) + 5O₂(g) ⟶ 4NO(g) + 6H₂O(g)

62,8 g of NH₃ are:

62,8g×(1mol/17,031g) = <em>3,69 moles of NH₃</em>

62,8 g of O₂ are:

62,8g×(1mol/32g) =<em> 1,96 moles of O₂</em>

For a complete reaction of 1,96 moles of O₂ you need:

1,96mol O₂×(4mol NH₃ / 5molO₂) = 1,57 moles NH₃. As you have 3,69 moles, limiting reactant is O₂.

Assuming a complete reaction, 1,96mol O₂ produce:

1,96mol O₂×(6mol H₂O / 5molO₂) = 2,35 moles of H₂O. In grams:

2,35 moles of H₂O×(18,01g/1mol) = <em>42,3g of H₂O</em>

I hope it helps!

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If 16.0 grams of aluminum oxide were actually produced, what is the percent yield of the reaction below given that you start wit
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Answer: The percent of yield is 50.03%.

Explanation:

First, we need to balance the equation:

4Al + 3O_{2} ⇒ 2Al_{2}O_{3}

We need to remember that the chemical equations are written in moles, so we have to convert the amounts in grams to moles, using the molecular weight of every compound: Al2O3 (101.96 g/mol), Al (29.98 g/mol) and O2 (31.99 g/mol).

In consequence, the amounts in moles of every compound will be:

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19 g O_{2}*\frac{1 mol}{31.99 g} =0.594 mol O_{2}

Now, we have to find out which is the limit reagent or in other words, which of the reagents will be consumed first, taking into account the stoichiometric ratio of the balanced equation:

0.333 mol Al * \frac{2 mol Al_{2}O_{3}}{4 mol Al} =0.1665 mol Al_{2}O_{3}

0.594 mol O_{2} * \frac{2 mol Al_{2}O_{3}}{3 mol O_{2}} =0.396 mol Al_{2}O_{3}

As you can see, the maximum amount (theoretically) of Al2O3 that can be produced is 0.1665 mol.

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3 years ago
How many moles of cei3 are in 9.42 × 1021 formula units of cei3 (cerium(iii) iodide; a bright, water, soluble solid). answer in
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1 mol of any substance is made of 6.022 x 10²³ units.
When its 1 mol of a formula, its made of 6.022 x 10²³ formula units
Therefore if 6.022 x 10²³ formula units of CeI₃ make up 1 mol of CeI₃
Then 9.42 x 10²¹ formula units make up - 1/(6.022 x 10²³) x (9.42 x 10²¹) = 0.0156 mol of CeI₃
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