Đáp án:

 

Giải thích các bước giải:

2kclo3 -> kcl +3o2

nkclo3 = 24,5/122,5= 0.2

-> no2 = 0,2*3/2= 0,3

Vo2= 0.3*22,4 =6,72 l

B, vo2 + vh20 = 6,72 + 8,96 = 15,68

mh20 = 15,68/22,4 *18 = 12,6 g

Câu 2 :

A. Zn + h2so4 -> znso4 + h2

C% h2so4 = mct*100%/mdd -> mct = 15%*200/100% = 30g

nh2so4 = 30/98 = 0,3

nzn= 13/65= 0.2

Mol zn hết tính nh2 theo nzn

nh2 = 0.2* 22.4 = 4.48l

B. nh2so4 dư = 0.1

C% h2so4 dư = 0.1*98/213= 0.046

C% znso4 = 0.2 * 161/213= 0,15

C. Cuo + h2 -> cu +h20

ncuo = 12/80 = 0.15

mcu = 0.15* 64 = 9.6g

 

1/

$a,PTPƯ:2KClO_3\xrightarrow{t^o} 2KCl+3O_2$

$n_{KClO_3}=\dfrac{24,5}{122,5}=0,2mol.$

$Theo$ $pt:$ $n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,3mol.$

$⇒V_{O_2}=0,3.22,4=6,72l.$

$b,PTPƯ:2H_2+O_2\xrightarrow{t^o} 2H_2O$

$n_{H_2}=\dfrac{8,96}{22,4}=0,4mol.$

$\text{Lập tỉ lệ:}$ $\dfrac{0,4}{2}<\dfrac{0,3}{1}$

$⇒O_2$ $dư.$

$Theo$ $pt:$ $n_{H_2O}=n_{H_2}=0,4mol.$

$⇒m_{H_2O}=0,4.18=7,2g.$

2/

$a,PTPƯ:Zn+H_2SO_4\xrightarrow{} ZnSO_4+H_2↑$

$n_{Zn}=\dfrac{13}{65}=0,2mol.$

$m_{H_2SO_4}=200.15\%=30g.$

$⇒n_{H_2SO_4}=\dfrac{30}{98}=0,3mol.$

$\text{Lập tỉ lệ:}$ $\dfrac{0,2}{1}<\dfrac{0,3}{1}$

$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,2mol.$

$⇒V_{H_2}=0,2.22,4=4,48l.$

$b,n_{H_2SO_4}(dư)=0,3-\dfrac{0,2.1}{1}=0,1mol.$

$⇒C\%_{H_2SO_4}=\dfrac{0,1.98}{13+200-0,2.2}.100\%=4,6\%$

$Theo$ $pt:$ $n_{ZnSO_4}=n_{Zn}=0,2mol.$

$⇒C\%_{ZnSO_4}=\dfrac{0,2.161}{13+200-0,2.2}.100\%=15,09\%$

$c,PTPƯ:CuO+H_2\xrightarrow{t^o} Cu+H_2O$

$n_{CuO}=\dfrac{12}{80}=0,15mol.$

$\text{Lập tỉ lệ:}$ $\dfrac{0,15}{1}<\dfrac{0,2}{1}$

$⇒H_2$ $dư.$

$Theo$ $pt:$ $n_{Cu}=n_{CuO}=0,15mol.$

$⇒m_{Cu}=0,15.64=9,6g.$

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