a)PTHH:\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b)\(n_{O_2}=\dfrac{3,36}{22,4}=0,162\left(mol\right)\Rightarrow m_{O_2}=0,162.32=5,184\left(g\right)\)
Theo PTHH ta có:\(n_{Al}=\dfrac{4}{3}n_{O_2}=\dfrac{4}{3}.0,162=0,216\left(mol\right)\Rightarrow m_{Al}=0,216.27=5,832\left(g\right)\)
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a) PTHH:
4 Al + 3 O2-->to-->2 Al2O3
b)nO2 = 3,3622 * 4=0,162(mol)
=> mO2 = 0,162 * 32 = 5,184(g)
TPTHH ta có nAl = 43 nO2 = 43 * 0,162 = 0,216(mol)
=> mAl = 0,216 * 27 = 5,832(g)
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