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Stolb23 [73]
3 years ago
10

When 12 moles of methanol (ch3oh) and 24 moles of oxygen gas react according to the chemical equation below, what is the limitin

g reactant and how much of the excess reactant is left over?
Chemistry
1 answer:
Nataliya [291]3 years ago
5 0
The balanced reaction equation for the reaction between CH₃OH and O₂ is 
                        2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l)
Initial moles        12                24
Reacted moles   12                18
Final moles          -                   6              12               24
 
The stoichiometric ratio between CH₃OH and O₂ is 2 : 3
Hence,
   reacted moles of O₂ = reacted moles of CH₃OH x (3/2)
                                      = 12 mol x 3 / 2  
                                      = 18 mol

All of CH₃OH moles react with O₂.

Hence, the limiting agent is CH₃OH. 

Excess reagent is O₂.
Amount of moles of excess reagent left = 24 - 18 mol = 6 mol
                                            
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An unknown insoluble substance displaced the water shown. It's mass is indicated on the triple beam balance. Mass = A. 694 B. 69
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Explanation:

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(b) Moles of H_2SO_4 can be calculated as:

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Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

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The conversion of mL to L is shown below:

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Thus, volume = 10×10⁻³ L

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1 mole of H_2SO_4 react with 2 moles of NaOH

0.00145 mole of H_2SO_4 react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 /  21.37×10⁻³  M = 0.1357 M

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