a chemist had a container of the compound kmno4 that weighted 790 grams. what was e weight of the oxygen in…
a chemist had a container of the compound kmno4 that weighted 790 grams. what was e weight of the oxygen in the container (k,39; mn, 55; o, 16)
Answer
Explanation:
Step1: Calculate molar mass of KMnO₄
$M(KMnO_4)=39 + 55+4\times16=158$ g/mol
Step2: Calculate moles of KMnO₄
$n(KMnO_4)=\frac{m(KMnO_4)}{M(KMnO_4)}=\frac{790}{158}=5$ mol
Step3: Calculate moles of oxygen
In 1 mole of KMnO₄, there are 4 moles of oxygen. So in 5 moles of KMnO₄, $n(O)=5\times4 = 20$ mol
Step4: Calculate mass of oxygen
$m(O)=n(O)\times M(O)=20\times16 = 320$ g
Answer:
320 g