Question

2.61 g of KClO3 were present in a 5.50 g samole of KClO3 -KCl mixture, what is the pecent by mass of KClO3 in the mixture?

also please show percent if the 2.61 were 0.904 thank you

Answer 1

2.61 g of KClO3 were present in a 5.50 g sample of KClO3 -KCl mixture, what is the percent by mass of KClO3 in the mixture?

also please show percent if the 2.61 were 0.904

$$\omega = \frac {m (KClO3)}{m (KClO3) + m (KCl)} * 100 \%$$ $$\omega = \frac {2.61}{5.50} * 100\% = 47.45\% wt$$ $$\mathrm{W(KClO3)} = 47.45\% \mathrm{wt}$$

if 2.61 of KClO3 = 90.4%

$$\mathrm{W} = 100\% - 90.4\% = 9.6\%$$ $$\mathrm{W} = 9.6\%$$