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Alex17521 [72]
3 years ago
10

Can anyone help me? I'm so confused.

Chemistry
2 answers:
Alenkasestr [34]3 years ago
7 0
Woah, that doesn't look like high school. That's gotta be college level.
nalin [4]3 years ago
5 0
Try changing the grade level on your answer maybe someone a little older would be better for helping with that good luck! :))
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Consider the process used to produce iron metal from its ore.
Dmitry_Shevchenko [17]

Answer:

223.4 grams of iron can be produced from 2.5 moles of Fe2O3 and 6.0 moles of CO.

Explanation:

The balanced reaction is:

Fe₂O₃ (s) + 3 CO(g) → 2 Fe(s) + 3 CO₂ (g)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

  • Fe₂O₃: 1 mole
  • CO: 3 moles
  • Fe: 2 moles
  • CO₂: 3 moles

Being:

  • Fe: 55.85 g/mole
  • O: 16 g/mole
  • C: 12 g/mole

the molar mass of the compounds participating in the reaction is:

  • Fe₂O₃: 2*55.85 g/mole + 3*16 g/mole= 159.7 g/mole
  • CO: 12 g/mole + 16 g/mole= 28 g/mole
  • Fe: 55.85 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole

Then, by stoichiometry of the reaction, the following quantities participate in the reaction:

  • Fe₂O₃: 1 mole* 159.7 g/mole= 159.7 g
  • CO: 3 moles* 28 g/mole= 84 g
  • Fe: 2 moles* 55.85 g/mole= 111.7 g
  • CO₂: 3 moles* 44 g/mole= 132 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

So, first of all, you can apply the following rule of three: if by reaction stoichiometry 1 mole of Fe₂O₃ reacts with 3 moles of CO, then 2.5 moles of Fe₂O₃ react with how many moles of CO?

moles of CO=\frac{2.5 moles of Fe_{2} O_{3}*3 moles of CO }{1 mole of Fe_{2} O_{3}}

moles of CO= 7.5

But 7.5 moles of CO are not available, 6.0 moles are available. Since you have less moles than you need to react with 2.5 moles of Fe₂O₃, CO will be the limiting reagent.

Now you can apply the following rule of three: if by reaction stoichiometry 3 moles of CO produce with 111.7 grams of Fe, then 6 moles of CO will produce how much mass of Fe?

mass of Fe=\frac{6 moles of CO*111.7 grams of Fe}{3 moles of CO}

mass of Fe= 223.4 grams

<u><em>223.4 grams of iron can be produced from 2.5 moles of Fe2O3 and 6.0 moles of CO.</em></u>

5 0
2 years ago
If you weighed out 0.38 g of calcium and reacted it with an excess of hcl, how many moles of h2(g) would you expect to produce?
Hitman42 [59]
Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
Chemical reaction: Ca + 2HCl → CaCl₂ + H₂.
m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.
4 0
3 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
a cell on a microscope slide and noticed cell walls and cell membrane bound organelles. the students conclude the cells are what
lord [1]
The students conclude the cells are plant cells.
7 0
3 years ago
If you are supposed to put in 2/4 a cup sugar for a recipe, can you just put in a half cup of sugar?
natta225 [31]
I would assume so.

Given \frac{2}{4}, we can simplify the fraction to \frac{1}{2}

Both would obtain the same proportions, so I don't see why putting a half cup of sugar would make things any different.

Hope this is the answer you are looking for.
8 0
3 years ago
Read 2 more answers
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