Câu hỏi của Nguyễn Thanh Hùng

Question

Câu1:Cho 11,2gam Fe tác dụng với dung dịch HCl Tính

A,thể tích khí H2 thu được ở đktc.(ĐS:4,48 lít)

B,khối lượng HCl phản ứng.(ĐS:14,6 g)

C,khối lượng FeC...

B.Thị Anh Thơ · Accepted Answer

$PTHH:Fe+2HCl\rightarrow FeCl_2+H_2$

________0,2____0,4______ 0,2_____0,2

a, Ta có:

$n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)$

$\Rightarrow V_{H2}=0,2.22,4=4,48\left(l\right)$

b,

$m_{HCl}=0,4.\left(1+35,5\right)=14,6\left(g\right)$

c,

$m_{FeCl2}=0,2.\left(56+35,5.2\right)=25,4\left(g\right)$

Câu trả lời:

$PTHH:Fe+2HCl\rightarrow FeCl_2+H_2$

________0,2____0,4______ 0,2_____0,2

a, Ta có:

$n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)$

$\Rightarrow V_{H2}=0,2.22,4=4,48\left(l\right)$

b,

$m_{HCl}=0,4.\left(1+35,5\right)=14,6\left(g\right)$

c,

$m_{FeCl2}=0,2.\left(56+35,5.2\right)=25,4\left(g\right)$

Thảo Phương · Answer

Fe+2HCl---------> FeCl2 + H2

$n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)$

a) Theo PT: $n_{H_2}=n_{Fe}=0,2\left(mol\right)$

=> V_H2= 0,2. 22,4 = 4,48(l)

b) Theo PT: $n_{HCl}=2n_{Fe}=0,4\left(mol\right)$

=> m_HCl= 0,4. 36,5 = 14,6 (g)

c) Theo PT: $n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)$

=> m_FeCl2= 0,2. 127 = 25,4 (g)

Câu trả lời:

Fe+2HCl---------> FeCl2 + H2

$n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)$

a) Theo PT: $n_{H_2}=n_{Fe}=0,2\left(mol\right)$

=> V_H2= 0,2. 22,4 = 4,48(l)

b) Theo PT: $n_{HCl}=2n_{Fe}=0,4\left(mol\right)$

=> m_HCl= 0,4. 36,5 = 14,6 (g)

c) Theo PT: $n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)$

=> m_FeCl2= 0,2. 127 = 25,4 (g)