Question

Please Help Me!!!!!!!!

2Na + Cl2 -> 2NaCl
If you want to make 82.59g of NaCl how many grams of each reactant are you going to need?

If you have 25,82g of sodium metal, how liters of chloride are you going to use?

Answer 1

How much sodium chlorida should be produced?

How much of the excess reactant will be left over?

In the above problem actually 82,7g NaCl was produced what is the precent yield?

Answer 2

okay, for the first 3 questions you need to apply stoichiometry using the amount of moles, of either the reactants or products, and the coefficients.

Set up a ratio using the species of interest, like so:

e.g. for a general reaction:

\(\color{red}{a}A + \color{blue}{b}B\) \(\rightleftharpoons\) \( \color{green}{c}C\)

where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients ,

\(\dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\)

From here you can isolate what you need.
For example: if you have 2 moles of B, how many moles of C can you produce?

solve algebraically: \(\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)
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To interconvert between mass and moles, use the relationship:

\(n=\dfrac{m}{M}\)

where, M=molar mass, m=mass, and n= moles.

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Answer 3

@aaronq so a: 2 b: 1 c: 2, than the m: is the mass of each elemt is that correct?

Answer 4

A: Na, B: Cl2 or Cl? C: NaCl?

Answer 5

B would be \(Cl_2\).

Answer 6

the coefficients are also right.

Answer 7

\[\frac{ 22.9Na }{ 2 } =\frac{ 70.9Cl2 }{ 1 }=\frac{ 58.4NaCl }{ 2 }\] @aaronq like this?

Answer 8

no, not the molar mass, n = moles

so for example, for the first question:

If you want to make 82.59g of NaCl how many grams of each reactant are you going to need?

you first need to convert the mass to moles, so \(n=\dfrac{82.59g}{58.4g/mol}=~1.4 moles\)

Finding the moles of \(Cl_2\)

\(\dfrac{1.4~moles}{2}=\dfrac{n_{Cl_2}}{1}\rightarrow n_{Cl_2}=\dfrac{1.4~moles*1}{2}=0.7~moles\)

now convert back to mass, \(0.7~moles=\dfrac{m}{70.9g/mol}\rightarrow m=100.19 g\)

Answer 9

So for the second one be a ration of 2:1 or turning the gram to litter? @aaronq im sorry to bother you chemistry is my worst subject :/ i get lost

Answer 10

for the second, you need to find the moles of \(Cl_2\) needed to react with amount of Na, yes, the ratio will be 2:1.

Once you find the moles of \(Cl_2\) you need to convert it to volume. If it's at STP you can multiply it by 22.414 L/mol. If it's at any other temp/pressure you need to use the ideal gas law, PV=nRT.

Answer 11

1. (82.59 g NaCl) / (58.4430 g NaCl/mol) x (2 mol Na / 2 mol NaCl) x (22.9898 g Na/mol) = 32.49 g Na
(82.59 g NaCl) - (32.49 g Na) = 50.10 g Cl2

2. (25.82 g Na) / (22.9898 g Na/mol) x (1 mol Cl2 / 2 mol Na) x (22.414 L/mol) = 12.59 L Cl2 at STP

@aarong like this?

Answer 12

yeah, thats right.

Answer 13

if you mix 59g of sodium with 59 grams chlorine gas what is the limiting reactant and excess ? @aaronq im sorry i am horrible in chemistry :/

Answer 14

convert to moles and divide by their respective stoichiometric coefficient.

Whichever has less (after the division) is the limiting reactant.

Answer 15

59g x 1 mol/ 22.990 = 2.566 / 2 [ from 2Na in equation] --> 1.28 59 g x 1 mol/ 71 = .83 / 1 ---> .83, like that? @aaronq

Answer 16

yep thats it!

Answer 17

@aaronq thanks, how do i find how much chloride should be produce? would that be .83 ? and how can i tell how much of the excess will remain? thanks again !

Answer 18

you use the exact same process you've used in the past questions to find the amount of NaCl you can make with the limiting reactant.

Subtract the moles to find how much there is left.

Answer 19

so Na (23) + Cl2( 71) = 58.49, 82.59/ 58.49 = 1.41?
or the 1.28 - .83 = .45 excess left?
@aaronq SORRy for all this

Answer 20

1.28 -0.83 = 0.45 moles of Na

But this is without taking the stoichiometric coefficient into account.

you need to multiply it by it, so 2*0.45 moles

Answer 21

So excess reactant will be .9 moles of Na?
1.41 or .83 of sodium choride will be produced?
@aaronq

Answer 22

the excess is right.

not really sure i follow the other part.

Answer 23

The question as " How much sodium chloride should be produced?" @aaronq

Answer 24

okay, so use a ratio: \(\dfrac{n_{Cl_2}}{1}=\dfrac{n_{NaCl}}{2}\)

Answer 25

\[\frac{ 71cl2 }{ 1 } = \frac{ 58NaCl }{ 2 }\]? @aaronq

Answer 26

nope, remember n means moles not molar mass.

\(\dfrac{0.83~moles}{1}=\dfrac{n_{NaCl}}{2}\)

Answer 27

so 4.14/ 2? @aaronq

Answer 28

what is that?

Answer 29

would it be 150.345 g of sodium chlorine would be produced? @aaronq

Answer 30

so there were 1.66 moles of NaCl

and therefore 1.66 moles*58.44 g/mol =97.0104 g