6.88 grams of [tex]Na_{2} CO_{3}[/tex] is produced from 11 grams of NaHC[tex]O_{3}[/tex].
Explanation:
Balance chemical reaction for the process:
2NaHC[tex]O_{3}[/tex] ⇒ [tex]Na_{2}[/tex][tex]CO_{3}[/tex] + [tex]H_{2}[/tex]O + C[tex]O_{2}[/tex]
Data given:
mass of NaHC[tex]O_{3}[/tex] given = 11 gram
atomic mass of NaHC[tex]O_{3}[/tex] = 84.007 grams/mole
mass of [tex]Na_{2}[/tex][tex]CO_{3}[/tex] produced = ?
number of moles of NaHC[tex]O_{3}[/tex] will be calculated as:
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
putting the values in the equation:
number of moles = [tex]\frac{11}{84.007}[/tex]
= 0.13 moles
from the balanced equation it can be seen that,
2 moles of NaHC[tex]O_{3}[/tex] reacted to give 1 mole of [tex]Na_{2} CO_{3}[/tex]
0.13 moles of NaHC[tex]O_{3}[/tex] will give x moles of [tex]Na_{2} CO_{3}[/tex]
[tex]\frac{1}{2}[/tex] = [tex]\frac{x}{0.13}[/tex]
x = [tex]\frac{0.13}{2}[/tex]
= 0.065 moles of [tex]Na_{2} CO_{3}[/tex]
Atomic mass of [tex]Na_{2} CO_3}[/tex] = 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[tex]CO_{3}[/tex] 6.88 grams of [tex]Na_{2} CO_{3}[/tex] is produced.
