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ki77a [65]
3 years ago
11

For the reaction ?H2 +?O2 → ?H2O, what is the maximum amount of H2O (18.0153 g/mol) which could be formed from 9.43 g of H2 (2.0

1588 g/mol) and 12.98 g of O2 (31.9988 g/mol)? Answer in units of g.
Chemistry
1 answer:
Montano1993 [528]3 years ago
3 0
2H2 + O2 →2H2O
This is a limiting reactant problem in which you need to figure out which reactant is the limiting reactant. You can usually look at the information given and see that hydrogen has the smaller amount of mass used, but it is always good to check that assumption through stoichiometry:
(9.43g H2/2.02g/mol)(2mol H2O/2mol H2)(18.02gH2O/1mol H2O)= 84.1g H2O produced from 9.43g of H2

(12.98g O2/32.00g/mol O2)(2mol H2O/1mol O2)(18.02gH2O/1mol H2O)= 14.6g H2O

So my assumption that hydrogen makes the smaller amount of water is wrong since oxygen (it is the limiting reactant) created the smaller amount of water with 14.6g, so 14.6g is the maximum amount of water that can be formed
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A reaction mixture initially contains 0.140 MCO and 0.140 MH2O. What will be the equilibrium concentration of CO?
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Answer:

0.013 M

Explanation:

From the question, we can make the following deductions; we are given mixture that contains two compounds, that is A and B, 0.140 M CO and 0.140 M H2O respectively. Then, we are asked to find the equilibrium concentration of Carbonmonoxide,CO.

So, the equation for the reaction is given below;

CO + H2O <-----------------> CO2 + H2.

Initially: we have 0.14M of CO, 0.14M of H2O and zero (0) concentration of CO2 and H2.

At time,t = CO =0.14 - x , H2O = 0.14 - x, CO2 and H2 = x.

The above reaction consist of the forward reaction and the backward reaction.

Therefore, the equilibrium Concentration of CO;

(Since we are giving that Kc = 102). Then, Kc=  [CO2][H2] ÷ [CO][H2O]. Where Kc is the equilibrium constant.

Therefore, 102 = [x^2] / [0.14 - x]^2.

==> 10.1= x/0.14 - x.

====> 0.141 - 10.1 x = x.

x + 10.1 x = 0.141.

===> 11.1 x = 0.141.

===> x = 0.141 ÷ 11.1.

===> x = 0.127 M .

Then, at time,t CO = 0.14 - x.

= 0.14 - 0.127 = 0.013 M

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