Chemistry (Tamil Medium) - StudyGuideIndia.com

G.Lô.2
10 kg çV KClO 3
ElûT ØÝYÕUôL, £ûRdÏm úTôÕ GqY[Ü ¨û\
Bd³_u YôÙ ¡ûPdÏm GuTûR LQd¡ÓL (AÔ¨û\ : K = 39, O = 16,
Cl = 35.5)
2KClO 3
→ 2 KCl + 3O 2
KClO 3
«u êXdáß ¨û\ = 39 + 35.5 + 48 = 122.5
O 2
²u êXdáß ¨û\ = 16 + 16 = 32
N¬NUu ùNnVlThP NUuTôh¥u T¥, (2 × 122.5) × 10 -3 Kg KCIO 3
ûV
ùYlTlTÓj§]ôp (3 × 32) × 10 -3 kg O 2
¡ûPd¡\Õ, G]úY
10kg KClO 3
RÚYÕ =
−3
3× 32×
10
2 × 122.5 × 10
−3
× 10 = 3.92 kg of O 2
G.Lô.3
200 ¡¡ ¨û\Ùs[ 90% çnûUVô] CaCO 3
ûY ùYlTlTÓjÕm ùTôÝÕ
¡ûPdÏm ÑhP ÑiQôm©u (CaO) ¨û\ Gu]?
CaCO 3
→ CaO + CO 2
100 × 10 -3 kg 56 × 10 -3 kg
200 kg 90% çnûUVô] CaCO 3
Cu ¨û\ = 90 200
100 ×
= 180 kg çV CaCO 3
100 × 10 -3 kg çV CaCO 3
ãÓTÓjÕm ùTôÝÕ 56 × 10 -3 kg CaO
¡ûPd¡\Õ. G]úY
U VW
180 kg CaCO 3
ûV ãÓTÓjÕm =
ùTôÝÕ ¡ûPdÏm CO 2
× ×
100 × 10
−3
56 10 180
−3
= 100.8 kg CaO
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