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3 years ago

H2O2 → H2O + O2

Chemistry

1 answer:

vladimir2022 [97]3 years ago

8 0

Answer:

75.47

Explanation:

AS WE CANT WEIGHT AIR THE ANSWER IS CORRECT

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Sodium metal and chlorine gas are combined in a reaction to produce sodium chloride also known as table salt. What are the react

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Give formula for iron (III) oxide

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Fe2O3

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3 years ago

Copper (II) Sulfate pentahydrate, CuSO4 X 5H20 is heated in an open crucible to remove the water. (a) Diagram the lab set up cle

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The water molecules are not completely removed so additional heating is required.

Explanation:

We have the copper (II) sulfate pentehydrate with the chemical formula CuSO₄ · 5H₂O.

molar mass of CuSO₄ · 5H₂O = 159.6 + 5 × 18 = 249.6 g/mole

Knowing this, we devise the following reasoning:

if in 249.6 g of CuSO₄ · 5H₂O there are 90 g of H₂O

then in 8 g of CuSO₄ · 5H₂O there are Y g of H₂O

Y = (8 × 90) / 249.6 = 2.88 g of water

mass of dried CuSO₄ = mass of CuSO₄ · 5H₂O - mass of H₂O

mass of dried CuSO₄ = 8 - 2.88 = 5.12 g

5.12 g is less that the weighted mass of 6.50 g. We deduce from this that the sample needs additional heating in order to remove all the water (H₂O) molecules.

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3 years ago

In Haber’s process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, wha

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Answer:

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

Step 1: Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate limiting reactant

For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

Step 4: Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

Step 5: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

Step 6: Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

Step 7: The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

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