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denis23 [38]
3 years ago
14

Drag each tile to the correct location.

Chemistry
2 answers:
RideAnS [48]3 years ago
7 0

  2 Fe2O3  + 3 C→  4 Fe  + 3 CO3

 The reactant  and  products in this  chemical   equation is as below


<u><em>Reactant </em></u>                  

Fe2O3

 C

<u>Product</u>

Fe

CO2


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

A chemical   reaction  involve  rearrangement  reactants   to  form  products.

Reactants are  written  on the  left  side   of  a chemical reaction  while  products are  written  on the  right hand  side  of chemical reaction.


Since Fe2O3  and C are  on  left side  of   chemical  reaction  they are the reactants.

Fe  and CO2 are  on the  right side  of  chemical  reaction they are  the products.




Anastasy [175]3 years ago
4 0

Here's the answer ⬇⬇⬇⬇

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4 years ago
Question 1:
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Question 1:

(a) Sulfurous acid: H2SO3

Sulfuric acid: H2SO4

(b) Nitrous acid: H2NO2

Nitric acid: H2NO3

Question 2:

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5 0
3 years ago
For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e
sergiy2304 [10]

Answer:a)  11.34 g of ethane (C_2H_6) can be formed

b) C_2H_4 is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

1. \text{Moles of} H_2=\frac{4.21}{2}=2.10moles

2. \text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles

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According to stoichiometry :

1 mole of C_2H_4 require 1 mole of H_2

Thus 0.378 moles of C_2H_4 will require=\frac{1}{1}\times 0.378=0.378moles  of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product and H_2 is the excess reagent.

moles of H_2 left = (2.10-0.378) = 1.72 moles

mass of H_2 left=moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g

According to stoichiometry :

As 1 mole of C_2H_4 give = 1 mole of C_2H_6

Thus 0.378 moles of C_2H_4 give =\frac{1}{1}\times 0.378=0.378moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g

Thus 11.34 g of ethane is formed.

4 0
4 years ago
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