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qaws [65]
3 years ago
12

The reaction of sodium peroxide and water produces sodium hydroxide and oxygen gas. The following balanced chemical equation rep

resents the reaction.
2 Na^2O^2(s) + 2 H2O(l) → 4 NaOH(s) + O^2(g)

If 15.7 moles of sodium hydroxide are produced, how many moles of O2 will be made?
Chemistry
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

3.925 mol.

Explanation:

  • From the balanced equation:

<em>2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g) ,</em>

It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂ .

<em>Using cross multiplication:</em>

4 moles of NaOH produced with → 1 mole of O₂ .

15.7 moles of NaOH produced with → ??? mole of O₂ .

<em>∴ The no. of moles of O₂ made =</em> (1 mole)(15.7 mole)/(4 mole) =  <em>3.925 mol.</em>

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Question is chemistry:
Black_prince [1.1K]

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Formula: H2S

Molar mass: 34.1 g/mol

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Melting point: -115.6°F (-82°C)

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8 0
3 years ago
Compounds X and Y are both C6H13Cl compounds formed in the radical chlorination of 3-methylpentane. Both X and Y undergo base-pr
zubka84 [21]

Answer:

Y is a 3-chloro-3-methylpentane.

The structure is shown in the figure attached.

Explanation:

The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).

The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.

6 0
2 years ago
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia (NH_3)

5 0
2 years ago
A chemical reaction produces 138 mol of HBr gas. What volume will that gas occupy at STP?
stepan [7]

Answer:

The second one

Explanation:

5 0
2 years ago
Read 2 more answers
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