How many grams of sodium oxide can be produced when 55.3 g Na react with 64.3 g O2? Unbalanced equation: Na + O2 → Na2O

Show, or explain, all of your work along with the final answer.

Answers

Answer 1
Answer:

Answer : The amount of sodium oxide produced is, 74.40 grams

Solution : Given,

Mass of sodium = 55.3 g

Mass of O_2= 64.3 g

Molar mass of sodium = 23 g/mole

Molar mass of O_2 = 32 g/mole

Molar  mass of Na_2O = 61.9 g/mole

First we have to calculate the moles of sodium and oxygen.

Moles of Na = \frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= (55.3g)/(23g/mole)=2.404moles

Moles of O_2 = \frac{\text{ given mass of }O_2}{\text{ molar mass of }O_2}= (64.3g)/(32g/mole)=2.009moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

4Na+O_2\rightarrow 2Na_2O

From the balanced reaction we conclude that

4 moles of sodium react with 1 mole of oxygen

2.404 moles of sodium react with (2.404)/(4)=0.601 moles of oxygen

Excess moles of oxygen = 2.009 - 0.601 = 1.408 moles

That means sodium is a limiting reagent and oxygen is an excess reagent.

Now we have to calculate the moles of sodium oxide.

From the reaction we conclude that,

4 moles of sodium react to give 2 moles of sodium oxide

2.404 moles of sodium react to give (2)/(4)* 2.404=1.202 moles of sodium oxide

Now we have to calculate the mass of sodium oxide.

\text{Mass of }Na_2O=\text{Moles of }Na_2O* \text{Molar mass of }Na_2O

\text{Mass of }Na_2O=(1.202moles)* (61.9g/mole)=74.40g

Therefore, the amount of sodium oxide produced is, 74.40 grams

Answer 2
Answer: Given the two amounts of reactants, we get the limiting reactant between the two. We convert each to mole and divide them to their corresponding stoichiometric coefficient. we balance first above, that is 4Na+O2=2Na2O. hence, for Na that is 0.60 mole and oxygen is 2.00 mole. We follow the smaller amount, Na, hence the amount of sodium oxide in grams is 74.53 grams. 

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Answers

She needs to use the heat of fusion (ΔH(fusion)) and the equation q=mΔH(fusion) or q=nΔH(fusion)  depending on whether or not ΔH(fusion) is given in J/g or J/mol.  The reason why temperature is absent from the equation is that temperature does not change at a phase change, only the enthalpy does.

I hope his helps.  Let me know in the comments if anything is unclear.

Answer:

Hfus

Explanation:

This is correct on edge.

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Answers

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Suppose that a substance in a beaker is heated over a burner in a science lab. Which observation would most likely indicate that a chemical change has occurred in the substance?1) If the substance is a liquid or solid, an increase in temperature would indicate a chemical change.

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3) If the substance is a solid, a change of some of the solid to liquid form would indicate a chemical change.

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Answers

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In an energy diagram, the reactants are at a higher potential energy compared to the products. Which of the following best describes this reaction?A)

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B)

It is endothermic and will have a negative enthalpy

C)

It is exothermic and will have a positive enthalpy.

D)

It is exothermic and will have a negative enthalpy.

Answers

Answer is (D) - It is exothermic and will have a negative enthalpy.


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What is the theoretical yield of H2O if 130 g of H2O is produced from 18 g of H2 and an excess of O2?

Answers

The theoretical yield of H2O if 130g of H2O is produced from 18g of H2 and an excess of O2 is 160g. 

Answer: The theoretical yield of water is 162 grams.

Explanation:

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

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

Given mass of hydrogen gas = 18 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrogen gas}=(18g)/(2g/mol)=9mol

The chemical equation for the formation of water from hydrogen gas and oxygen gas follows:

2H_2+O_2\rightarrow 2H_2O

As, oxygen gas is present in excess. It is considered as an excess reagent.

So, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrogen gas produces 2 moles of water

So, 9 moles of hydrogen gas will produce = (2)/(2)* 9=9mol of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 9 moles

Putting values in equation 1, we get:

9mol=\frac{\text{Mass of water}}{18g/mol}\n\n\text{Mass of water}=162g

Hence, the theoretical yield of water is 162 grams.