All categories

3 years ago

What is the mass of 8.83 x 10^23 formula units of Iron (III) oxide, Fe2O3?

Chemistry

1 answer:

zepelin [54]3 years ago

8 0

Answer:

234.68g

Explanation:

We understood from Avogadro's hypothesis that 1mole of any substance contains 6.02x10^23 atoms. Therefore, 1mole of Fe2O3 also contains 6.02x10^23 atoms

1 mole of Fe2O3 = (56x2) + (16x3) = 112 + 48 = 160g

Now, if 160g of Fe2O3 contains 6.02x10^23 atoms,

Therefore Xg of Fe2O3 will contain 8.83×10^23 atoms i.e

Xg of Fe2O3 = (160x8.83×10^23)/6.02x10^23 = 234.68g

You might be interested in

Suggest why helium can be used as a carrier gas but oxygen cannot.

SCORPION-xisa [38]

Because helium is lighter then the oxygen .and due to this helium gass carrier go upward easily .

7 0

3 years ago

Read 2 more answers

A science teacher asked students to calculate the density of a toy block. The mass of the block was 110 grams and the volume of

Kipish [7]

Answer:

13.75

Explanation:

Density is equal to mass divided by volume. 110/8 = 13.75

5 0

3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

3 years ago

What is the molar solubility of agcl in 0. 30 m nacl at 25°C. ksp for agcl is 1. 77 × 10^-10.

Rzqust [24]

Molar solubility of AgCl will be 0.59 × $10^{-10}$ M.

The amount of a chemical that can dissolve in one liter of a solution before reaching saturation is known as its molar solubility. This implies that the quantity of a substance it can disintegrate in a solution even before the solution becomes saturated with that particular substance is determined by its molar solubility.

A compound's molar solubility would be the measure of how many moles of such a compound must dissolve to produce one liter of saturated solution. The molar solubility unit will be mol L-1.

Calculation of molar solubility:

Given data:

M = 0.30 M

$K_{sp}$ = 1.77 × $10^{-10}$