Question

How much Na2Co3 is produced from 11g of NaHCo3

Answer 1

6.88 grams of
`Na_(2) CO_(3)` is produced from 11 grams of NaHC
`O_(3)`.

Step-by-step explanation:

Balance chemical reaction for the process:

2NaHC
`O_(3)` ⇒
`Na_(2)`
`CO_(3)` +
`H_(2)`O + C
`O_(2)`

Data given:

mass of NaHC
`O_(3)` given = 11 gram

atomic mass of NaHC
`O_(3)` = 84.007 grams/mole

mass of
`Na_(2)`
`CO_(3)` produced = ?

number of moles of NaHC
`O_(3)` will be calculated as:

number of moles =
`(mass)/(atomic mass of 1 mole)`

putting the values in the equation:

number of moles =
`(11)/(84.007)`

= 0.13 moles

from the balanced equation it can be seen that,

2 moles of NaHC
`O_(3)` reacted to give 1 mole of
`Na_(2) CO_(3)`

0.13 moles of NaHC
`O_(3)` will give x moles of
`Na_(2) CO_(3)`

`(1)/(2)` =
`(x)/(0.13)`

x =
`(0.13)/(2)`

= 0.065 moles of
`Na_(2) CO_(3)`

Atomic mass of
`Na_(2) CO_3}` = 105.98 grams/mole

mass = atomic mass x number of moles

mass = 0.065 x 105.98

mass = 6.88 grams

so, from 11 grams of NaH
`CO_(3)` 6.88 grams of
`Na_(2) CO_(3)` is produced.