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Rom4ik [11]
3 years ago
9

When methane is burned with oxygen, the products are carbon dioxide and water, If you produce 20 grams of water and 10 grams of

carbon dioxide from 15 grams of oxygen, how many grams of methane were needed for the reaction?​
Chemistry
2 answers:
serious [3.7K]3 years ago
5 0

Answer:

4 g

Explanation:

First you need to write a balanced chemical equation. You are given thatmethane is burned, meaning a combustion reaction in which carbon dioxide and water are released.

Unbalanced: CH4 + O2 ---> CO2 + H2O

Balanced: CH4 + 2O2 ---> CO2 + 2H2O

Givens:

X grams CH4 (Molecular mass 16.0 grams)

9 grams H2O (Molecular mass 18.0 grams)

11 grams CO2 (Molecular mass 44.0 grams)

Mole ratio: 1:2:1:2 (CH4:O2:CO2:H2O)

Then you need to find which of the reactants are the limiting reactant and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.

11 g CO2 / (44.0 g) = 0.25 moles CO2

9 g H2O / (18.0 g) = 0.5 moles

n of CH4 = n of CO2 = n of H2O /2 = 0.25 moles

m of CH4 = n* Mw = 0.25 * 16.0 = 4 g

scZoUnD [109]3 years ago
3 0
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction? First you need to write a balanced chemical equation.
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6210 (0.1050)

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2 years ago
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4 0
3 years ago
The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t
slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<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)

Given mass of iron(III) phosphate = 20.00 g

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Putting values in equation 1, we get:

\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol

The given chemical equation follows:

2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = \frac{1}{2}\times 0.133=0.0665moles of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0
3 years ago
Your theoretical yield is 81.2 grams, and your actual yield is 78.2 grams. What is the percent yield?
ioda

Answer:

66.2

Explanation:

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2 years ago
Which of the following are the bark, roots, seeds, buds or berries of an aromatic plant?
djverab [1.8K]

Answer:

Spices

Explanation:

Herbs are the leaves.

Vegetables are the seeds.

Fruits are the seeds.

So spices are the only option left.

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2 years ago
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