Question

C4H5O5 + 3NaHCO3 -> C4H2Na3O5+ 3H2O + 3CO2

Information:
n(NaHCO3)= 5,94 mol
Maleic acid (MW: 134 g/mol)

used 1.1 times more of the substance quantity of the acid than the base.

How much malic acid (C4H5O5) to be used? Show calculation; give the result in g

Answer 1

We have acid C4H5O5, and base NaHCO3, It says that we use 1.1 times more acid. we'll use 1.1 later.

Answer 2

you have not got his @simoneforver ur trying it from yesterday right

Answer 3

sam may help you out

Answer 4

Yes I hope so :)

Answer 5

@.Sam. common answer the @simoneforver question and burn it like anything

Answer 6

So, we can find out the number of moles of acid, n(C4H5O5) using the moles of n(NaHCO3) by just dividing it,

According to
C4H5O5 + 3NaHCO3 -> C4H2Na3O5+ 3H2O + 3CO2

3 NaHCO3 gives 1 C4H5O5 , so,
\[5.94\cancel{molNaHCO_3} \times \frac {1 molC_4H_5O_5}{3\cancel{mol NaHCO_3}}=1.98molC_4H_5O_5\]

Answer 7

Aa ok then the malic acid in gram is m=1,98g *134g/mol = 265,32g

Answer 8

No first I have to do this... 1,98mol *1,1 right?

Answer 9

Yes and

Answer 10

Yes so it is m=(1,98*1,1)mol*134g/mol =

Answer 11

\[2.178mol \times \frac{134g}{1mol}=\]

Answer 12

Yes :) thanks for your help..

Answer 13

welcome :)