Question

cho 9,2g Na tac dung voi 1,6g khi O2 thu duoc natri oxit (Na2O) xac dinh cac chat sau phan ung (biet H=1;O=16;Cl=35,5;Mg=24;S=32

Answer 1

\(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{1,6}{32}=0,05\left(mol\right)\\ 4Na+O_2\rightarrow2Na_2O\\ Vì:\dfrac{0,4}{4}>\dfrac{0,05}{1}\Rightarrow Na.dư\\ Sau.phản.ứng:Na_2O,Na\left(dư\right)\\ n_{Na\left(dư\right)}=0,4-4.0,05=0,2\left(mol\right)\\ m_{Na\left(dư\right)}=0,2.23=4,6\left(g\right)\\ n_{Na_2O}=2.0,05=0,1\left(mol\right)\\ m_{Na_2O}=0,1.62=6,2\left(g\right)\)