$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
$n_{KClO_3} = \dfrac{24,5}{122,5} = 0,2(mol)$
$n_{O_2} = \dfrac{3}{2}n_{KClO_3} = 0,3(mol)$
$V_{O_2} = 0,3.22,4 = 6,72(lít)

Đáp án B

\(KClO_3-^{t^o}\rightarrow KCl+\dfrac{3}{2}O_2\)

\(n_{O_2}=\dfrac{3}{2}n_{KClO_3}=\dfrac{3}{2}.\dfrac{24,5}{122,5}=0,3\left(mol\right)\)

=> \(V_{O_2}=0,3.22,4=6,72\left(l\right)\)

=> Chọn B

Đáp án B

n K C l O 3 = 12,25 122,5 = 0,1   m o l

2 K C l O 3   →   2 K C l   +   3 O 2 0 , 1                     →                                   0 , 15           m o l

V O 2 =   0 , 15 . 22 , 4   =   3 , 36   l

\(n_{HCl}=\dfrac{146.50}{100.36,5}=2\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

_______________2----------------->1

=> V_CO2 = 1.22,4 = 22,4 (l)

=> C

Mọi người giải luôn hộ nhé

Bài 1:

\(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\\ PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\\ \left(mol\right).....0,1\rightarrow....0,1.......0,15\\ a,m_{KCl}=0,1.74,5=7,45\left(g\right)\\ V_{O_2}=0,15.22,5=3,36\left(l\right)\)

CaCO3 bạn nhé

Bài 2:

\(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\\ PTHH:CaCO_3\underrightarrow{t^o}CaO+CO_2\\ \left(mol\right).....0,1\rightarrow....0,1.....0,1\\ m_{CaO}=0,1.56=5,6\left(g\right)\\ V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

Zn+2HCl->Zncl2+H2

0,1-------------------0,1

n Zn=0,1 mol

=>VH2=0,1.22,4=2,24l

=>C

C

C

a) 

\(n_{SO_2}=\dfrac{m}{M}=\dfrac{3,2}{64}=0,05\left(mol\right)\\ n_{CO_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

b) 

\(n_{Cl_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\Rightarrow m_{Cl_2}=n.M=0,06.71=4,26\left(mol\right)\\ n_{Na_2CO_3}=n.M=0,5.106=53\left(g\right)\)

c) 

\(V_{N_2\left(\text{đ}ktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\\ n_{O_2}=\dfrac{m}{M}=\dfrac{4,8}{32}=0,15\left(mol\right)\Rightarrow V_{O_2\left(\text{đ}ktc\right)}=n.22,4=0,15.22,4=3,36\left(l\right)\)