Bài 12:
a, PT: \(2Cu+O_2\underrightarrow{t^o}2CuO\)
\(2Mg+O_2\underrightarrow{t^o}2MgO\)
b, Vì số nguyên tử Cu gấp 2 lần số nguyên tử Mg
→ Số mol Cu gấp 2 lần số mol Mg.
Gọi: nCu = 2x (mol) ⇒ nMg = x (mol)
Theo PT: \(\left\{{}\begin{matrix}n_{CuO}=n_{Cu}=2x\left(mol\right)\\n_{MgO}=n_{Mg}=x\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow2x.80+40x=30\Rightarrow x=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,3\left(mol\right)\\n_{Mg}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m=m_{Cu}+m_{Mg}=0,3.64+0,15.24=22,8\left(g\right)\)
c, Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Cu}+\dfrac{1}{2}n_{Mg}=0,225\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,225.22,4=5,04\left(l\right)\)
\(\Rightarrow V_{kk}=\dfrac{5,04}{20\%}=25,2\left(l\right)\)
