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12345 [234]
3 years ago
5

The molar mass of H2O is 18.01 g/mol. The molar mass of O2 is 32.00 g/mol. What mass of H2O, ins grams, must react to produce 50

.00 g of O2

Chemistry
1 answer:
Ivan3 years ago
5 0

Answer:

56.28 g

Explanation:

First change the grams of oxygen to moles.

(50.00 g)/(32.00 g/mol) = 1.5625 mol O₂

You have to use stoichiometry for the next part.  Looking at the equation, you can see that for every 2 moles of H₂O, 1 mole of O₂ is produced.  Convert from moles of O₂ to moles of H₂O using this relation.

(1.5625 mol O₂) × (2 mol H₂O/1 mol O₂) = 3.125 mol H₂O

Now convert moles of H₂O to grams.

(3.125 mol) × (18.01 g/mol) = 56.28125 g

Convert to significant figures.

56.28125 ≈ 56.28

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We have to discuss here whether mole should be considered as a base unit in SI system or not.

The answer is-   mole is considered as base unit in SI system.

There are seven base units in the SI system. They are- kilogram (kg) for mass, second (s) for time, kelvin (K) for temperature, ampere (A) for electric current, mole (mol) for the amount of substance, candela (cd) for luminous intensity and meter (m) is for distance.

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How many moles of hydrogen gas would be needed to react with excess carbon dioxide to produce 30.6 moles of water vapor?
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If you have 3.8 grams of NaCl, how many moles of NaCl do you have?
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6 0
3 years ago
If 16.29 grams of Na2SO4 is mixed with 3.697 grams of C and allowed to react according to the balanced equation: Na2SO4(aq) + 4
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<u>Answer:</u> The limiting reagent in the given chemical reaction is carbon metal.

<u>Explanation:</u>

Excess reagent is defined as the reagent which is present in large amount in a chemical reaction.

Limiting reagent is defined as the reagent which is present in small amount in a chemical reaction. Formation of product depends on the limiting reagent.

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

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

  • <u>For sodium sulfate:</u>

Given mass of sodium sulfate = 16.29 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

\text{Moles of sodium sulfate}=\frac{16.29g}{142g/mol}=0.115mol

  • <u>For carbon:</u>

Given mass of carbon = 3.697 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon}=\frac{3.697g}{12g/mol}=0.31mol

For the given chemical reaction:

Na_2SO_4(aq.)+4C(s)\rightarrow Na_2S(s)+4CO(g)

By Stoichiometry of the reaction:

4 moles of carbon reacts with 1 mole of sodium sulfate

So, 0.31 moles of carbon will react with = \frac{1}{4}\times 0.31=0.0775mol of sodium sulfate

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

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

Hence, the limiting reagent in the given chemical reaction is carbon metal.

4 0
3 years ago
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