Question

cho 6 gam Mg td vs H2SO4. Hay cho biet the tich khi H2 sinh ra o dktc

Answer 1

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

\(n_{Mg}=\dfrac{m}{M}=\dfrac{6}{24}=0,25\left(mol\right)\)

Theo PTHH, \(n_{H_2}=n_{Mg}=0,25\left(mol\right)\)

\(V_{H_2}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\)

Answer 2

PTHH: Mg + H₂SO₄ \(\rightarrow\) MgSO₄ + H_{2\(\uparrow\)}
n_Mg = \(\dfrac{6}{24}\) =0,25(mol)
Theo PT: n_H2 = n_Mg = 0,25 (mol)
=> V_H2 = 0,25.22,4 =5,6 (l)

Answer 3

\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)
PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
Theo PT ta có: \(n_{Mg}=n_{H_2}=0,25\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)