\(n_{Cl_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(2NaBr+Cl_2\rightarrow2NaCl+Br_2\)
\(0.2............0.1\)
\(m_{NaBr}=0.2\cdot103=20.6\left(g\right)\)
Theo gt ta có: $n_{Cl_2}=0,1(mol)$
$Cl_2+2NaBr\rightarrow 2NaCl+Br_2$
Suy ra $n_{NaBr}=0,2(mol)\Rightarrow m_{NaBr}=20,6(g)$
Cl2+2NaBr ->2NaCl+Br2
0,1---0,2 mol
n Cl2=2,24\22,4=0,1 mol
=>m NaBr =0,2.103=20,6g
\(n_{Cl_2} = \dfrac{2,24}{22,4} = 0,1\ mol\\ 2NaBr + Cl_2 \to 2NaCl + Br_2\\ n_{NaBr} = 2n_{Cl_2} = 0,1.2 = 0,2(mol)\\ \Rightarrow m_{NaBr} = 0,2.103 = 20,6(gam)\)
