phan ung hoa hoc bi sai nhe
\(PTHH:2Ca+O_2->2CaO\)
0,2--->0,1---->0,2 (mol)
\(n_{Ca}=\dfrac{m}{M}=\dfrac{8}{40}=0,2\left(mol\right)\)
A
\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,1\cdot22,4=2,24\left(l\right)\)
B
C1
\(m_{CaO}=n\cdot M=0,2\cdot\left(40+16\right)=11,2\left(g\right)\)
C2
\(m_{O_2}=n\cdot M=0,1\cdot32=3,2\left(g\right)\)
ap dung DLBTKL ta co
\(m_{Ca}+m_{O_2}=m_{CaO}\\ =>8+3,2=m_{CaO}\\ =>m_{CaO}=11,2\left(g\right)\)