Answer:
= 14.24g of [tex](NH4)_{2}CO_{3}[/tex] is required.
Explanation:
Reaction equation:
[tex](NH4)_{2}CO_{3}[/tex] → [tex]2NH_{3} + CO_{2} + H_{2} O[/tex]
Mole ratio of ammonium carbonate to carbon dioxide is 1:1
1 mole of CO2 - 44g
?? mole of CO2 - 6.52g
= 6.52/44 = 0.148 moles was produced from this experiment.
Therefore, if 1 mole of [tex](NH4)_{2}CO_{3}[/tex] - 96.09 g
0.148 mol of [tex](NH4)_{2}CO_{3}[/tex] -- ?? g
=0.148 × 96.09
= 14.24g of [tex](NH4)_{2}CO_{3}[/tex] is required.
