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Luda [366]
3 years ago
14

When 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because:

Chemistry
1 answer:
noname [10]3 years ago
6 0

7.5 mol of hydrogen would be needed to consume the available nitrogen.

Explanation:

When hydrogen reacts with nitrogen, ammonia is formed as shown below;

3H₂ (g) + N₂ (g) → 2NH₃ (g)

As seen from the equation, every 3 moles of H₂ react with a mole of N₂ to form 2 moles of NH₃.

The limiting factor in a chemical reaction is the reactant that gets depleted first.

Because the molar mass of nitrogen gas is approximately 28g/mol, 70g of nitrogen gas would be 2.5 moles.

The reaction ratio of nitrogen to hydrogen in the reaction is 1 : 3. The reaction would require 2.5 * 3 (7.5) moles of hydrogen for a complete reaction.

However since there are only 7g on hydrogen, (Remember 1 mole of H₂ is approximately 2g), the available moles of H₂ is 7 / 2 = 3.5

3.5 moles fall short of the 7.5 moles of H₂ required for a complete reaction. H₂ gets depleted first before N₂.  The reaction would require 4 more moles of H₂.

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Hoochie [10]

Answer:

30.3 g

Explanation:

At STP, 1 mol of any gas will occupy 22.4 L.

With the information above in mind, we <u>calculate how many moles are there in 32.0 L</u>:

  • 32.0 L ÷ 22.4 L/mol = 1.43 mol

Then we <u>calculate how many moles would there be in 16.6 L</u>:

  • 16.6 L ÷ 22.4 L/mol = 0.741 mol

The <u>difference in moles is</u>:

  • 1.43 mol - 0.741 mol = 0.689 mol

Finally we <u>convert 0.689 moles of CO₂ into grams</u>, using its <em>molar mass</em>:

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6 0
3 years ago
How many grams of h2 will be produced if 175g of HCI are allowed to react completely with sodium
Sedbober [7]

Answer:

4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium

Explanation:

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:

  • HCl: 2 moles
  • Na: 1 mole
  • NaCl: 2 moles
  • H₂: 1 mole

You know the following masses of each element:

  • H: 1 g/mole
  • Cl: 35.45 g/mole
  • Na: 23 g/mole

So, the molar mass of each compound participating in the reaction is:

  • HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
  • Na: 23 g/mole
  • NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
  • H₂: 2* 1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:

  • HCl: 2 moles* 36.45 g/mole= 72.9 g
  • Na: 1 mole* 23 g/mole= 23 g
  • NaCl: 2 moles* 58.45 g/mole= 116.9 g
  • H₂: 1 mole* 2 g/mole= 2 g

So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

mass of H_{2} =\frac{175 g of HCl*2g ofH_{2} }{72.9 g of HCl}

mass of H₂= 4.8 g

<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>

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3 years ago
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Answer:

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C. Graph A

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What would be the major product obtained from hydroboration–oxidation of the following alkenes?
zmey [24]

Answer:

a. 3-methylbutan-2-ol

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Explanation:

For this reaction, we must remember that the hydroboration is an <u>"anti-Markovnikov" reaction</u>. This means that the "OH" will be added at the <em>least substituted carbon of the double bond.</em>

In the case of <u>2-methyl-2-butene</u>, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore <u>the "OH" will be added to carbon three</u> producing <u>3-methylbutan-2-ol</u>.

For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore<u> the "OH" will be added to carbon 2</u> producing <u>2-methylcyclohexan-1-ol</u>.

See figure 1

I hope it helps!

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