nKClO3=nKCl; m(KClO3)M(KClO3)=m(KCl)M(KCl){\frac {m(KClO3)} {M(KClO3)}=\frac {m(KCl)} {M(KCl)}}M(KClO3)m(KClO3)=M(KCl)m(KCl) m(KCl)=m(KClO3)∗M(KCl)M(KClO3)m(KCl)=\frac {m(KClO3) ∗ M(KCl)} {M(KClO3)}m(KCl)=M(KClO3)m(KClO3)∗M(KCl) ; m(KCl) = 2.433 (grams).
SOLUTION: 2.433 grams
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