Be prepared effervescent tablets. You use 500g sodium bicarbonate (MW: 84 g / mol). To make effervescent mixture will be malic acid (MW: 134 g / mol). To effervescent solution should not taste bad as soap with rather little tart, use 1.1 times more of the substance quantity of the acid than the base. a) Write shower action between malic acid and sodium bicarbonate. b) How much malic acid to be used? Show calculation; give the result in g
can you write the reaction equation between the malic acid and the sodium carbonate
3NaHCO3 + C4O5H5 -> .......3Co2 + 3H2O i really don't know
humnn try to balance it along with CO2 u should get COO-Na should come right side
C2H2(COOH)2 + Na2CO3 -> C2H2(COONa)2 + H2O + CO2
yaa now its correct
mean first one u alredy did right
which one the first one?
first option of the question
a) Write shower action between malic acid and sodium bicarbonate.
Yes :) Now i have to finde the answer for b)
that is confusing meet too
I have to finde how much malic acid has to be used i g... How do i do that?
The balanced equation of C2H2(COOH)3 + NaHCO3 -> ?
C4H5O5 + 3NaHCO3 -> C4H2Na3O5+ 3H2O + 3CO2
Now i can answer b)
yes please let me know
Do you know how to do it?
noo just cracking the head
How much malic acid to be used to what i want ? the question is not clear i find
To prepare effervescent tablets
We have a acid-base reaction C4H5O5 + 3NaHCO3 -> C4H2Na3O5+ 3H2O + 3CO2 x*1,1 5,94mol n(NaHCO3) = m/M = 500g/84 g/mol = 5,94mol x= m= n*M -> x*134 g/mol = ¬-------------g
how much we have to prepare the eff tablet
yes
not yes how much
Be prepared effervescent tablets. You use 500g sodium bicarbonate (MW: 84 g / mol). To make effervescent mixture will be malic acid (MW: 134 g / mol). To effervescent solution should not taste bad as soap with rather little tart, use 1.1 times more of the substance quantity of the acid than the base. b) How much malic acid to be used? Show calculation; give the result in g
To make it clear.. I have to finde out how much malic acid to be used i g.. And my answer is We have a acid-base reaction C4H5O5 + 3NaHCO3 -> C4H2Na3O5+ 3H2O + 3CO2 x*1,1 5,94mol n(NaHCO3) = m/M = 500g/84 g/mol = 5,94mol x= m= n*M -> x*134 g/mol = ¬-------------g But how do i found x?
iam not getting it :(
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