For The Reaction Shown Find The Limiting Reactant For Each Of The Following Initial Amounts Of

For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants.?

The equation tells you that 2molK react with 1 mol Cl2
Ratio = 2:1
a).1 mol K; 1 molCl2 : Ratio = 1:1 K is limiting
b).1.8 mol K; 1 molCl2 - Ratio 1.8:1 K is limiting
c).2.2 mol K; 1 molCl2 = Ratio 2.2:1 = Cl2 is limiting
d).14.6 mol K; 7.8 molCl2 Ratio 14.6:7.8 = 14.6/7.8 = 1.87:1 K is limiting

For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants. 2Al(s)+3Cl2(g)→2AlCl3(s)?

(1.0 g Al) / (26.98154 g Al/mol) = 0.03706 mol Al
(1.0 g Cl2) / (70.9064 g Cl2/mol) = 0.01410 mol Cl2
0.01410 mole of Cl2 would react completely with 0.01410 x (2/3) = 0.00940 mole of Al, but there is more Al present than that, so Al is in excess and Cl2 is the limiting reactant.

(2.2 g Al) / (26.98154 g Al/mol) = 0.08154 mol Al
(1.8 g Cl2) / (70.9064 g Cl2/mol) = 0.02539 mol Cl2
0.02539 mole of Cl2 would react completely with 0.02539 x (2/3) = 0.01693 mol of Al, but there is more Al present than that, so Al is in excess and Cl2 is the limiting reactant.

(0.353 g Al) / (26.98154 g Al/mol) = 0.013083 mol Al
(0.482 g Cl2) / (70.9064 g Cl2/mol) = 0.0067977 mol Cl2
0.0067977 mole of Cl2 would react completely with 0.0067977 x (2/3) = 0.0045318 mole of Al, but there is more Al present than that, so Al is in excess and Cl2 is the limiting reactant.

Find the limiting reactant for each of the following initial amounts of reactants.?

At Mass Al = 26
Mol Mass O2 = 32
Mol Mass Al2O3 = 100

Needed for reaction

4 moles Al
3 moles O2

to yield

2 moles Al2O3

The mole ratio of Al to O2 should be 4:3 = 1.33

a)
1.0g Al = 1/26 moles
1.0 g O2 = 1/32 moles
The ratio is 32/26 = 1.23, there is not enough Al, so Al is the limiting reactant

b)
2.2g Al = .0846 mole
1.8g O2 = .0563 mole
The ratio is 1.504, there is too much Al so O2 is the limiting reactant

c)
0.353g Al = 0.0136 mole
0.482g O2 = 0.0151 mole
The ratio is 0.901, not enough Al, so Al is the limiting reactant

For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants.?

shall we initiate off with one. If there's a million mole of Cr then by employing molar ratios O2 mmust be 3/4 moles. Howeve there are a million moles of 02 relatively. subsequently the O2 is in extra and the cr is the proscribing. enable's see the different. If there are a million mole of 02 then there might desire to be 4/3 moles of cr. even with the undeniable fact that there is in basic terms a million moles of cr and subsequently the cr is proscribing the reaction. was hoping this help, in case you like greater help. Mail me.

Help finding limiting reactant?

For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants.

2 Na(s) + Br2(g) ---> 2NaBr(s)

(i) 2mol Na, 2 mol Br2
Express your answer as a chemical formula.


(ii) 1.8mol Na, 1.4 mol Br1
Express your answer as a chemical formula.


(iii) 2.5mol Na, 1 mol Br2
Express your answer as a chemical formula.


(iv) 12.6mol Na, 6.9 mol Br2
Express your answer as a chemical formula.

For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants.?

10.5/(2* 6.94) = 0.76 < 37.2/38=.98
========> Li is limiter

Chemistry help: finding limiting reactants?

2Al(s) + 3S(g)→Al2S3(s)
2 mol 3 mol
From the above reaction it is evident that 2 mol of Al reacts completely with 3 mol of S.
Hence 1 mole of Al will require 3/2 i.e 1.5 mol of S for complete reaction.
Thus if the mol of S is less than 1.5 times that of Al, S will be limiting reactant. On the other hand if the mol of S is greater than 1.5 times that of Al, Al will be limiting reactant. In case there is 1.5 times mol of S to that of Al, the reaction will be complete of course theoretically.

Part A:
1.0 g Al = 1.0 g/27 g mol-1 = 0.037 mol
1.0 g S = 1.0 g/32 g mol-1 = 0.03125 mol
Here the mol of S is less than 1.5 mol of Al, so S is limiting reactant.

Part B:
2.2 g Al = 2.2 g/ 27 g mol-1 = 0.08148 mol
1.8 g S = 1.8 g/32 g mol-1 = 0.05625 mol
The same as in Part A. S is limiting reactant

Part C:
0.353 g Al = 0.353 g/27 g mol-1 = 0.01307
0.482 g S = 0.482 g/32 g mol-1 = 0.01506
The same as above. S is limiting reactant.

For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants.?

Chemical equation: 4Cr (s) +3O2 (g) ----> 2Cr2O3 (s)

The balanced equation above states that 4 moles of Cr react with 3 moles of O2 to produce 2 moles of Cr2O3

So, 1 mole of Cr needs 0.75 moles of O2.
If the mole ratio of Cr : O2 is not 1 : 1.333, then you have an excess of one reagent. The other reagent is the limiting reagent.

Set up a proportion to determine the amount of O2 needed.

The dots are to prevent yahoo answers from scrunching the letters and numbers to the left!!

.. (moles of Cr) ……… 1
.. ___________ .. = .. ____
.. (moles of O2) …… 0.75


Problem 1:

1 mol Cr ; 1 mol O2
1 mole of Cr needs 0.75 moles of O2, you have 1 mole of O2, so you have excess O2, so Cr is the limiting reagent!

Problem 2:

4 mol Cr ; 2.5 mol O2

.. (4 moles of Cr) ……… 1
.. ___________ ... = .. ____
.. (moles of O2) ..…… 0.75

(4 * 0.75) ÷ 1 = 3 moles of O2 needed

4 moles of Cr need 3 moles of O2, you only have 2.5 mole of O2, so O2 is the limiting reagent.

Problem 3:
12 mol Cr ; 10 mol O2

.. (12 moles of Cr) ……… 1
.. ___________ ….. = .. ____
.. (moles of O2) ….…… 0.75

(12 * 0.75) ÷ 1 = 9 moles of O2 needed

12 moles of Cr need 9 moles of O2, you have 10 moles of O2, so you have excess O2, so Cr is the limiting reagent!





Problem 4:

14.8 mol Cr; 10.3 mol O2

.. (14.8 moles of Cr) ……… 1
.. ___________ …... = .. ____
.. (moles of O2) …..…… 0.75

(14.8 * 0.75) ÷ 1 = 11.1 moles of O2 needed

14.8 moles of Cr need 11.1 moles of O2, you only have 10.3 moles of O2, so O2 is the limiting reagent.

FIND THE LIMITING REACTANT: PLEASE HELP!!!?

Limiting reactant means which has the least moles?
Whats a mol? A mol is a unit. 1 mol = 6.022*10^23 molecules.
In this question, they give you a balanced equation, and then tell you how many grams of each you get. Convert from grams to moles, and see which reactant creates the least moles of product. Find molar mass on a periodic table to convert from grams to moles.
To calculate:
A grams Al * (1mol Al /26.982g Al) * ( 2 mol Al2O3/4mol Al) = B mol Al2O3
X grams O2 * (1mol O2 /15.99g Al) * ( 2 mol Al2O3/3mol O2) = Y mol Al2O3
Whichever option creates less moles of Al2O3 is the limiting reagent. Eg. if B is less than Y, than Al is the limiting reagent. All you have to do now is sub in the number of grams for A and X in the above equations respectively