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ozzi
4 years ago
13

How many moles of SnCl2 will be produced if 85.3 g of FeCl3 is reacted with an excess of Sn for the following reaction:2 FeCl3(s

) + 3 Sn(s) → 3 SnCl2(s) + 2 Fe(s)
Chemistry
1 answer:
Ronch [10]4 years ago
4 0

Answer:

The answer to your question is 0.79 moles of SnCl₂  

Explanation:

Data

moles of SnCl₂ = ?

mass of FeCl₃ = 85.3 g

excess Sn

Balanced chemical reaction

                 2 FeCl₃  +  3 Sn  ⇒   3 SnCl₂  +  2 Fe

Process

1.- Convert the mass of FeCl₃ to moles

Molar mass of FeCl₃ = 56 + (35.5 x 3)

                                  = 56 + 106.5

                                  = 162.5 g

Use proportions to find the moles of FeCl₃

                       162.5 g -------------------- 1 mol

                         85.3 g -------------------  x

                         x = (85.3 x 1) / 162.5

                         x = 0.525 moles

2.- Find the number of moles SnCl₂

                     2 moles of FeCl₃ ----------------- 3 moles of SnCl₂

                     0.525 moles        ----------------- x

                     x = (0.525 x 3) / 2

                    x = 0.79 moles of SnCl₂                  

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Answer:

2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)

Entalpy=-2861.9~KJ

Explanation:

In this case, we have to start with the <u>reagents</u>:

Al~+~NH_4NO_3

The compounds given by the problem are:

-) <u>Nitrogen gas</u> =  N_2

-) <u>Water vapor</u>  =  H_2O

-) <u>Aluminum oxide</u> =  Al_2O_3

Now, we can put the products in the <u>reaction</u>:

Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_)

When we <u>balance</u> the reaction we will obtain:

2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)

Now, for the enthalpy change, we have to find the <u>standard enthalpy values</u>:

Al_(_S_)=0~KJ/mol

NH_4NO_3_(_a_q_)=-132.0~KJ/mol

N_2_(_g_)=0~KJ/mol

H_2O_(_g_)=~-~241.8~KJ/mol

Al_2O_3_(_S_)=~-~1675.7~KJ/mol

With this in mind, if we <u>multiply</u> the number of moles (in the balanced reaction) by the standard enthalpy value,  we can calculate the energy of the <u>reagents</u>:

(0*2)~+~(-132*3)=~-396~KJ

And the <u>products</u>:

(0*3)~+~(-241.8*6)~+~(-1675.7*1)=-3125.9~KJ

Finally, for the total enthalpy we have to <u>subtract</u> products by reagents :

(-3125.9~KJ)-(-396~KJ)=-2729.9~KJ

I hope it helps!

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