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                2 LiOH +  CO2 à Li2CO3 +  H2O,


how much H2O will be produced if 25.4 g LiOH reacts with 64.8 g CO2?

Jul 11th, 2015

I'm so glad I could help you!

The trick to this problem is they are giving you two amounts of reactants and you have to determine which is the limiting reactant. A limiting reactant is the chemical that is used up first. Once that chemical is used up, the reaction is over.


In this problem our first step is to determine the limiting reactant. To do so, you convert the masses of reactants given in the problem to moles of reactant. When there is a coefficient in front of one of the terms, you must divide the number of moles by that number. (THINK ABOUT IT: If Joe has 10 cookies and I have 6, and we each eat one at a time, I will finish first. But if I eat one at a time and Joe eats 5 at a time, Joe is going to finish first.) The coefficient is simply the amount of reactant consumed at a time. Let's convert those masses into moles using molar masses.

Molar Mass LiOH: 23.95g/mol

25.4g LiOH is 1.06 mol of LiOH, but 2 will be consumed at once, so the reaction will be able to take place .53 times

Molar Mass CO2: 44.01g/mol

64.8g CO2 is 1.47 mol of CO2. The coefficient is one so based on the amount of CO2 the reaction would be able to occur 1.47 times, but since there is only enough LiOH for the reaction to take place .53 times the reaction will take place .53 times.


Now since we know how much reaction is occurring, we can determine the amount of water produced. We can use some stoichiometry in this case to convert 1.06 mol of LiOH to mass of H2O. 

1.06 mol LiOH x (1 mol H2O/2 mol LiOH) x 18.02g H2O/mol H2O = 9.55g H2O.

(THINK ABOUT IT: For every 2 cookies joe eats he gives 1 to tammy. Joe has 1.06 cookies tammy will end up with .53 cookies and we can convert that to the weight of cookies using molar mass and then even further to volume if we had the density of the cookie.)

The question asks how much. That answer can be moles, volume, or mass unless the problem specifies.

If you have any additional questions feel free to ask. No question is stupid, but not knowing the answer is stupid because you can just ask a question to learn the answer! ;) If I made a silly calculation error, feel free to let me know if it is causing you troubles.
Jul 11th, 2015

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