Đáp án:
\(b.\, a=1,2\\c.\, V_{\text{kk}}=2,8\text{(lít)}\\ d.\, m_{\text{KMn}O_4}=7,9\text{(gam)}\)
Giải thích các bước giải:
\(a.\, Mg+\dfrac 12O_2\xrightarrow{t^{\circ}} MgO\\ b. n_{Mg}=n_{MgO}=\dfrac 2{40}=0,05\text{(mol)}\to a=0,05\times 24=1,2\text{(gam)}\\ c. \, n_{O_2}=0,05\times \dfrac 12=0,025\text{(mol)}\to V_{O_2}=0,025\times 22,4= 0,56\text{(lít)}\to V_{kk}=\dfrac{0,56}{20\%}=2,8\text{(lít)}\\ d. \, 2KMnO_4\xrightarrow{t^{\circ}} K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4}=0,025\times 2=0,05\text{(mol)}\to m_{KMnO_4}=158\times 0,05=7,9\text{(gam)}\)