In the kinetic theory the particles in a gas behavior:____

Blast furnaces extra pure iron from the Iron(IIl)oxide in iron ore in a two step sequence. In the first step, carbon and oxygen

OLga [1]

Answer:

5.9 kg

Explanation:

We must work backwards from the second step to work out the mass of oxygen.

1. Second step

Mᵣ: 55.84

Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

m/kg: 7.0

(a) Moles of Fe

$\text{Moles of FeO} = \text{7000 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.84 g Fe}} = \text{125 mol Fe}$

(b) Moles of CO

$\text{Moles of CO} = \text{125 mol Fe} \times \dfrac{\text{3 mol CO}}{\text{2 mol Fe}} = \text{188 mol CO}$

However, this is the theoretical yield.

The actual yield is 72. %.

We need more CO and Fe₂O₃ to get the theoretical yield of Fe.

(c) Percent yield

$\begin{array}{rcl}\text{Percent yield} &=& \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \, \%\\\\ 72. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.72 &= &\dfrac{\text{188 mol}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{188 mol}}{0.72}\\& = & \textbf{261 mol}\\\\\end{array}$

We must use 261 mol of CO to get 7.0 kg of Fe.

2. First step

Mᵣ: 32.00

2C + O₂ ⟶ 2CO

n/mol: 261

(a) Moles of O₂

$\text{Moles of O}_{2} = \text{261 mol CO} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol CO}} = \text{131 mol O}_{2}$

(b) Mass of O₂

$\text{Mass of O}_{2}= \text{131 mol O }_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \text{4180 g O}_{2}$

However, this is the theoretical yield.

The actual yield is 71. %.

We need more C and O₂ to get the theoretical yield of CO.

(c) Percent yield

$\begin{array}{rcl}71. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.71 &= &\dfrac{\text{4180 g}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{4180 g}}{0.71}\\\\& = & \text{5900 g}\\& = & \textbf{5.9 kg}\\\end{array}$

We need 5.9 kg of O₂ to produce 7.0 kg of Fe.

6 0

4 years ago

In which type of bond are electrons shared equally between two atoms?

Natali5045456 [20]

Answer:

Nonpolar Covalent Bond

Explanation:

A nonpolar covalent bond is one in which the bonding electrons are evenly distributed between the two atoms.

Good Luck!

Hope this helps!

:)

6 0

3 years ago

Help I will give brainiest

Nataly [62]

Answer:

The first theory states that the rings formed at the same time as the planet. Some particles of gas and dust that the planets are made of were too far away from the core of the planet and could not be squashed together by gravity. They remained behind to form the ring system.

Explanation:

7 0

3 years ago

How many moles of O2 are required to react with 2.4 mol of H2 ?

Leokris [45]

Answer:

1.2 moles

Explanation:

this is the balanced equation for the reaction of oxygen (O2) and hydrogen (H2), usually we don't write the 1 in front of O2

2H₂ + 10₂ → 2H₂O

the molar ratio of hydrogen to oxygen is 2 : 1

we are trying to react with 2.4 mol of H2 so the moles of O2 is half the number of moles of H2 = 2.4 ÷ 2 = 1.2 mol

another way to think of it:

2H₂ + 10₂

2 : 1

2.4 mol : x mol

to get from 2 to 2.4 multiply by 1.2, so do the same to the other side

1 × 1.2 = 1.2 mol

8 0

2 years ago

Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for

eduard

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

7 0

4 years ago

In the kinetic theory the particles in a gas behavior:_____.