Question

Để điều chế được 3,36 lít khí oxi (ở đktc), thì cần nung nóng bao nhiêu gam kali clorat (KClO3)?

Answer 1

\(n_{O2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)

\(PTHH:2KClO_3\rightarrow2KCl+3O_2\)

_________0,1_____________0,15______mol

\(\Rightarrow m_{KClO3}=0,1.122,5=12,25\left(g\right)\)

Answer 2

\(2KClO_3\rightarrow2KCl+3O_2\)

\(n_{O_2}=\frac{3.36}{22.4}=0.15\left(mol\right)\\ \Rightarrow n_{KClO_3}=\frac{2}{3}\cdot n_{O_2}=\frac{2}{3}\cdot0.15=0.1\left(mol\right)\\ \Rightarrow m_{KClO_3}=0.1\cdot122.5=12.25\left(g\right)\)

Answer 3

a)2KClO3−−>2KCl+3O2

0,1---------------0,1--------0,15 mol

nO2=3,36\22,4=0,15(mol)

mKClO3=0,1.122,5=12,25(g)

Answer 4

2KCLO3-->2KCl+3O2

\(n_{O2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{KCLO3}=\frac{2}{3}n_{O2}=0,1\left(mol\right)\)

\(m_{KCLO3}=0,1.122,5=12,25\left(g\right)\)