\(n_{Cl_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ Cl_2+2NaI\rightarrow2NaCl+I_2\\ n_{NaI}=0,2.0,6=0,12\left(mol\right)\\ Vì:\dfrac{0,12}{2}>\dfrac{0,05}{1}\\ \Rightarrow NaIdư\\ ddX:NaI,NaCl\\ NaI_{dư}+AgNO_3\rightarrow AgI\downarrow\left(vàng.đậm\right)+NaNO_3\\ NaCl+AgNO_3\rightarrow AgCl\downarrow\left(trắng\right)+NaNO_3\\ n_{NaI\left(p.ứ\right)}=n_{NaCl}=2.n_{Cl_2}=0,1\left(mol\right)\\ n_{NaI\left(dư\right)}=0,12-0,1=0,02\left(mol\right)\\ n_{AgCl}=n_{NaCl}+n_{NaI\left(dư\right)}=0,1+0,02=0,12\left(mol\right)\\ \Rightarrow m=m_{\downarrow}=m_{AgCl}=143,5.0,12=17,22\left(g\right)\)
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