Câu hỏi của Nguyen Thu Hue

Question

cho 5,6g Sat tac dung voi AxitClohidric 1M
a. Viet PTHH 
b. Tinh the tich khi Hidro thu duoc o dktc
c.Tinh the tich Axitclhidric can dung

Hải Anh · Accepted Answer

a, PT: $Fe+2HCl\rightarrow FeCl_2+H_2$

b, Ta có: $n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)$

Theo PT: $n_{H_2}=n_{Fe}=0,1\left(mol\right)$

$\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)$

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

$\Rightarrow V_{ddHCl}=\frac{0,2}{1}=0,2\left(l\right)$

Bạn tham khảo nhé!Câu trả lời: a, PT: $Fe+2HCl\rightarrow FeCl_2+H_2$

b, Ta có: $n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)$

Theo PT: $n_{H_2}=n_{Fe}=0,1\left(mol\right)$

$\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)$

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

$\Rightarrow V_{ddHCl}=\frac{0,2}{1}=0,2\left(l\right)$

Bạn tham khảo nhé!

𝓓𝓾𝔂 𝓐𝓷𝓱 · Answer

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

b) Ta có: $n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)$

$\Rightarrow n_{H_2}=0,1mol$ $\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)$

c) Theo PTTH: $n_{HCl}=2n_{Fe}=0,2mol$

$\Rightarrow V_{HCl}=\frac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)$Câu trả lời: a) PTHH:  $Fe+2HCl\rightarrow FeCl_2+H_2$

b) Ta có: $n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)$

$\Rightarrow n_{H_2}=0,1mol$ $\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)$

c) Theo PTTH: $n_{HCl}=2n_{Fe}=0,2mol$

$\Rightarrow V_{HCl}=\frac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)$

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