Bài 3:
a)
\(n_{NaOH}=0,25.40=10\left(g\right)\)
b) \(m_A=\dfrac{1,2395}{24,79}.44+\dfrac{3,7185}{24,79}.64=11,8\left(g\right)\)
Bài 4:
\(n_{CO_2}=n_{CH_4}=n_{O_2}=n_{N_2}=n_{Cl_2}=\dfrac{6,1975}{24,79}=0,25\left(mol\right)\)
=> Phải lấy:
\(m_{CO_2}=0,25.44=11\left(g\right)\\ m_{CH_4}=0,25.16=4\left(g\right)\\ m_{O_2}=0,25.32=8\left(g\right)\\ m_{N_2}=0,25.28=7\left(g\right)\\ m_{Cl_2}=0,25.71=17,75\left(g\right)\)