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maks197457 [2]
2 years ago
15

How many moles of alumminum os needed to react woth 6.34 moles of fe3o4 in thereaction below? 8al + 3fe3o4_ 4al2o3 +9 fe

Chemistry
1 answer:
Alecsey [184]2 years ago
7 0

 The moles of Al  that is needed  to react with  6.34   moles of Fe3O4  is  16.9 moles


<u><em> Explanation</em></u>

<em>  </em>8 Al+ 3Fe3O4 →  4 Al2O3 + 9Fe


use of  mole  ratio  of Al: fe3O4 to calculate  the  moles of  Al

The mole  ratio  of  Fe3O4 : Al  is 8:3  therefore the  moles of  Al =

6.34 moles x8/3 = 16.9 moles

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Perform the following
Dmitry [639]

Answer:

0.5474

Explanation:

3.22 x 0.17

     \/

322 x 17 = 5,474

[] There are four places "after" the decimal in 3.22 and 0.17 combined, meaning we move the decimal to the right four times in our current answer

-> 5,474 becomes 0.5474

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

4 0
2 years ago
You have 125.0 mL of a solution of H3PO4, bu you don't know its concentration. if you titrate the solution with a 4.56-M solutio
irakobra [83]

Answer:

4.90 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = ? M

V₁ = 125.0 mL

M₂ = 4.56 M

V₂ = 134.1 mL

Using the above formula , the molarity of acid , can be calculated as ,

M₁V₁ = M₂V₂  

Substituting the respective values ,  

M₁ *  125.0 mL = 4.56 M *  134.1 mL

M₁ = 4.90 M

6 0
3 years ago
3<br>Nitrogen obtained in the laboratory is collected over water.why?​
Cloud [144]
Because displacement of water is the convenient way to obtain gas.
8 0
3 years ago
Balance the following equation. Then determine the ratio for the products KCl and O2 generated during the decomposition of potas
drek231 [11]
The answer is 2:3 for plato users

6 0
2 years ago
Read 2 more answers
What is the volume of 2 mol of chlorine gas at STP?<br> 2.0 L<br> 11.2 L<br> 22.4 L<br> 44.8 L
nirvana33 [79]

Answer:

44.8 L

Explanation:

Using the ideal gas law equation:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

At Standard temperature and pressure (STP);

P = 1 atm

T = 273K

Hence, when n = 2moles, the volume of the gas is:

Using PV = nRT

1 × V = 2 × 0.0821 × 273

V = 44.83

V = 44.8 L

7 0
2 years ago
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