Question #208422
Question text
Consider the following: KClO3 (122.55 g/mol), is prepared from KCl (74.55 g/mol) and O2 gas (32.00 g/mol) by the following process:
KCl + O2 -------------> KClO3 (unbalanced reaction)
The amount of oxygen gas in grams that will be needed to produce 14.2 mL of KClO3 (ρ = 2.34 g/mL) is
Expert's answer
Amount of oxygen gas in grams=34.2g
