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Anna71 [15]
3 years ago
7

15 POINTS FOR ANYONE WHO CAN ANSWER THIS!!!!!! plZ

Chemistry
1 answer:
yaroslaw [1]3 years ago
8 0

Answer:

A. 2Al(s) + 3Cl2(g) —> 2AlCl3(s)

B. 2Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

C. H2(g) + Br2(l) —> 2HBr(g)

Explanation:

A. Reaction between aluminium metal and chlorine gas.

Al(s) + Cl2(g) —> AlCl3(s)

The above equation can be balance as follow:

There are 2 atoms of Cl on the left side and 3 atoms on the right side. It can be balance by putting 3 in front of Cl2 and 2 in front of AlCl3 as shown below:

Al(s) + 3Cl2(g) —> 2AlCl3(s)

There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Al as shown below:

2Al(s) + 3Cl2(g) —> 2AlCl3(s)

Now the equation is balanced.

B. Reaction between lithium metal and liquid water.

Li(s) + H2O(l) —> LiOH(aq) + H2(g)

The above equation can be balance as follow:

There are 2 atoms of H on the left side and a total of 3 atoms on the right side. It can be balance by putting 2 in front of H2O and 2 in front of LiOH as shown below:

Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

There are 2 atoms of Li on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Li as shown below:

2Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

Now, the equation is balanced.

C. Reaction between gaseous hydrogen and liquid bromine.

H2(g) + Br2(l) —> HBr(g)

The above equation can be balance as follow:

There are 2 atoms of H on the left side and 1 atom on the right. It can be balance by putting 2 in front of HBr as shown below:

H2(g) + Br2(l) —> 2HBr(g)

Now the equation is balanced.

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Chemical equation:

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Balanced chemical equation:

2C₂H₆ +  7O₂     →      4CO₂ + 6H₂O

Step 1:

2C₂H₆ +  O₂     →      CO₂ + H₂O

Left hand side                      Right hand side

C = 4                                     C = 1

H = 12                                    H = 2

O = 2                                     O = 3

Step 2:

2C₂H₆ +  O₂     →      4CO₂ + H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 2

O = 2                                     O = 9

Step 3:

2C₂H₆ +  O₂     →      4CO₂ + 6H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 12

O = 2                                     O = 14

Step 4:

2C₂H₆ +  7O₂     →      4CO₂ + 6H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 12

O = 14                                     O = 14

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