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const2013 [10]
3 years ago
9

Molar mass CO2 is 44.01 g/mol. How many moles of LiOH are needed to react completely with 25.5 g of CO2?

Chemistry
2 answers:
poizon [28]3 years ago
7 0

Given:

Molar mass CO2 is 44.01 g/mol

25.5 g of CO2

 

Required:

Moles of LiOH

 

Solution:

Balanced equation is:

2LiOH +CO2 → Li2CO3 + H2O

 

25.5g CO2 (1 mole CO2/44.01 g/mol CO2)(2 moles LiOH/1 mol CO2) = 1.16 moles LiOH

joja [24]3 years ago
5 0

Answer:

1.16 moles of LiOH.

Explanation:

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Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ
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To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows 
time         0         40        80       120       160
moles    0.100   0.067  0.045    0.030    0.020


Q1)
for the first 40 s change of moles ;
      = -d[A] / t
      = - (0.067-0.100)/40s
      = 8.25 x 10⁻⁴ mol/s
for the next 40 s
      = -(0.045-0.067)/40
      = 5.5 x 10⁻⁴ mol/s
the 40 s after that
      = -(0.030-0.045)/40 s 
     = 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ   -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation. 
Since this is a first order reaction,
b = 1
therefore the reaction is 
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
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