How many grams of H2O are produced?

Question

LiOH + CO2 → Li2CO3 + H2O







If the percent yield of the above reaction is found to be 91.72 %, how many grams of H2O are produced from the complete reaction of 119.75 g of LiOH ?

Answer 1

First you have to balence the equation.

2LiOH + CO2 → Li2CO3 + H2O

119.75 g LiOH x 1 mol LiOH x 1 mol H2O x 18.015 g H2O
````````````````````` ---------------- -------------- ---------------
``````````````````````23.948 g LiOH 2 mol LiOH 1 mol H2O
= 45.04 g H2O (this is the theoretical yeild)

to get % yeild you take auctual yeild (what you're looking for) and divide it up theoretical yeild (the above answer), then to turn it into a % you multiply it by 100. So % yeild= (auctual/theoretical)x100.

simple algebra gets auctual by itself. (%yeild x theoretical)/100

91.72 x 45.04
------------------ = 41.31 g H2O
100

Answer 2

2LiOH + CO2 → Li2CO3 + H2O


Take the produced amount of

119.75 g LiOH x 23.95 g/mol LIOHx 1 mol H2O/2 Mol LiOh x18.0153 g/mol H20

that gives you the amount of water produced( theoretical amount) then you use the formula for percent yeild

percent yeild =(Acutal amount / theoretical amount) x 100%
and rearange it to get

actual amount = 91.72%/(100% x thoretical amount )

Answer 3

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