Câu 6.
\(n_P=\dfrac{12,4}{31}=0,4mol\)
\(n_{O_2}=\dfrac{33,6}{32}=1,05mol\)
\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
\(\dfrac{0,4}{4}\)< \(\dfrac{1,05}{5}\) ( mol )
0,4 0,5 0,2 ( mol )
Chất dư là O2
\(m_{O_2\left(dư\right)}=\left(1,05-0,5\right).32=17,6g\)
\(m_{P_2O_5}=0,2.142=28,4g\)
Câu 7.\(1m^3=1000l\)
\(n_{CH_4}=\dfrac{1000}{22,4}.98\%=43,75mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
43,75 87,5 ( mol )
\(V_{O_2}=87,5.22,4=1960l\)
Câu 8.
Gọi kim loại đó là R
\(4R+3O_2\rightarrow\left(t^o\right)2R_2O_3\)
\(n_{R_2O_3}=\dfrac{10,2}{2M_R+48}\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(4R+3O_2\rightarrow\left(t^o\right)2R_2O_3\)
\(\dfrac{30,6}{4M_R+96}\) <-- \(\dfrac{10,2}{2M_R+48}\) ( mol )
Ta có:
\(\dfrac{30,6}{4M_R+96}=0,15\)
\(\Leftrightarrow0,6M_R+14,4=30,6\)
\(\Leftrightarrow M_R=27\) ( g/mol )
=> R là Nhôm (Al)
