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3 years ago

Help!: Determine the mass of oxygen that can be removed from 25.0 grams of Fe2O3?

Chemistry

1 answer:

Lorico [155]3 years ago

3 0

7.5g

Explanation:

Given parameters:

Mass of Fe₂O₃ = 25g

Unknown:

Mass of oxygen that can be removed = ?

Solution:

Find the ratio of the oxygen atoms to that of the compound using the molar mass. Then multiply with the given mass.

Mass of oxygen = $\frac{molar mass of oxgygen}{molar mass of [tex]Fe_{2}O_{3}$ }[/tex] x given mass

Mass of oxygen = $\frac{3 x 16}{160}$ x 25 = 7.5g

Learn more:

Percentage composition brainly.com/question/8654167

#learnwithBrainly

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3 years ago

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3 years ago

Which one of the following statements is not true concerning 2.00 L of a 0.100 M solution of Ca3(PO4)2?

Len [333]

Answer: The correct answer is Option B.

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .......(1)

For A:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 2.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{2.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 2.00L)=0.200mol$

Moles of calcium phosphate = 0.200 moles

For B:

1 mole of calcium phosphate contains 3 moles of calcium atoms, 2 moles of phosphate atoms and 8 moles of oxygen atoms.

So, 0.200 moles of calcium phosphate will contain = $(8\times 0.200)=1.6$ moles of oxygen atoms.

Moles of oxygen atoms = 1.6 moles

For C:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 1.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{1.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 1.00L)=0.100mol$

Moles of calcium ions = $(0.100\times 3)=0.300$ moles

For D:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 5.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{5.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 5.00L)=0.500mol$

Moles of phosphorus atoms = $(0.500\times 2)=1.00$ moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of atoms

Number of phosphorus atoms in 0.500 moles of calcium phosphate will be = $(1.00\times 6.022\times 10^{23})=6.022\times 10^{23}$

For E:

1 mole of calcium phosphate contains 3 moles of calcium ions and 2 moles of phosphate ions.

So, 0.200 moles of calcium phosphate will contain = $(3\times 0.200)=0.600$ moles of calcium ions

Moles of calcium ions = 0.600 moles

Hence, the correct answer is Option B.

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3 years ago