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If 40g of Sodium (Na) reacts with 60g of Chlorine (Cl2), what is the mass of the products?

a. 40 g
b. 60 g
c. 160 g
d. 100 g


Sagot :

Answer:

Explanation:2

Na

(

s

)

+

Cl

2

(

g

)

2

NaCl

(

s

)

and that the reaction produced  

234 g

of sodium chloride. Convert this to moles by using the compound's molar mass

234

g

1 mole NaCl

58.44

g

=

4.004 moles NaCl

Now, you know that the sample of sodium metal reacted completely, which implies that it was completely consumed by the reaction.

In other words, you don't have to worry about the sample of chlorine gas because the fact that sample of sodium metal was completely consumed lets you know that the chlorine gas is not a limiting reagent.

This means that the reaction consumed

4.004

moles NaCl

2 moles Na

2

moles NaCl

=

4.004 moles Na

Convert this to grams by using the element's molar mass

4.004

moles Na

22.99 g

1

mole Na

=

92.1 g

−−−−−

 

The answer is rounded to three sig figs.

SIDE NOTE You can show that chlorine gas is not the limiting reagent by calculating the number of moles present in the sample

0.142

kg

1 mole Cl

2

35.453

kg

=

4.005 moles Cl

2

In order to produce  

4.004

moles of sodium chloride, you only need

4.004

moles NaCl

1 mole Cl

2

2

moles NaCl

=

2.002 moles Cl

2

Since you have more chlorine gas than you need, chlorine is in excess.

The mass of the products is d. 100 g

The required chemical reaction is given below

Na(s) + Cl₂(g) → 2NaCl(s)

From the chemical reaction, we see that 1 mole of Na reacts with 1 mole of Cl₂ to produce 2 moles of NaCl.

So, we need to find the limiting reagent in the reaction.

Since we have 40 g of Na, the number of moles of sodium = mass of sodium/molar mass of sodium = 40 g/23 gmol = 1.739 mole

Also, since we have 60 g of Cl₂, the number of moles of Chlorine = mass of chlorine/molar mass of chlorine = 60 g/(2 × 35.5 g/mol) = 60 g/71 g/mol = 0.845 mol

So, chlorine is the limiting reagent since it has less number of moles in the reaction.

Since from the chemical reaction, we require 1 mole of Cl to 2 moles of NaCl, the products. So, the number of moles of NaCl is 2 × 0.845 mole = 1.69 mol

Now mass of NaCl, the products = number of moles of NaCl × molar mass NaCl

= 1.69 mol × (23 g/mol + 35.5 g/mol)

= 1.69 × 58.5 g/mol

= 98.9 g

≅ 100 g

The mass of the products is d. 100 g

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