Question

dien phan hoan toan 20lit nuoc o trang thai long (D=1g/ml)voi hieu suat la 95%. The tich khi hidrothu dc o (dktc)la bao nhieu 

a, 25644,4lit                        b, 24644,4 lit 

c, 23644,4 lit                       d, 23640,4 lit 

Answer 1

PTHH: \(H_2O\xrightarrow[]{đp}H_2+\dfrac{1}{2}O_2\)

Ta có: \(m_{H_2O}=20000\cdot1=20000\left(g\right)\) \(\Rightarrow n_{H_2\left(lýthuyết\right)}=n_{H_2O}=\dfrac{20000}{18}=\dfrac{10000}{9}\left(mol\right)\)

\(\Rightarrow V_{H_2\left(thực\right)}=\dfrac{10000}{9}\cdot22,4\cdot95\%\approx23644,4\left(l\right)\)

\(\Rightarrow\) Đáp án C

 

 

Answer 2

Answer 3

\(V_{H_2O} = 20.1000.1 = 20 000(gam)\\ n_{H_2O} = \dfrac{20000}{18} = \dfrac{10 000}{9}(mol)\\ 2H_2O \xrightarrow{điện\ phân} 2H_2 + O_2 n_{H_2} = n_{H_2O\ pư} = \dfrac{10 000}{9}.95\% = \dfrac{9500}{9}(mol)\\ V_{H_2} = \dfrac{9500}{9}.22,4 = 23644,44(lít)\)

Đáp án C

Answer 4

PTHH: ⇒nH2(lýthuyết)=nH2O=2000018=100009(mol)⇒nH2(lýthuyết)=nH2O=2000018=100009(mol)