\(n_{NaBr}=0,8.0,5=0,4 \left(mol\right)\\ n_{Cl_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 2NaBr+Cl_2\rightarrow2NaCl+Br_2\\ Vì:\dfrac{0,4}{2}>\dfrac{0,15}{1}\Rightarrow NaBrdư\\ \Rightarrow n_{NaBr\left(dư\right)}=0,4-0,15.2=0,1\left(mol\right)\\ n_{NaCl}=2.0,15=0,3\left(mol\right)\\ AgNO_3+NaBr_{dư}\rightarrow NaNO_3+AgBr\downarrow\\ AgNO_3+NaCl\rightarrow AgCl\downarrow+NaNO_3\\ n_{AgBr}=n_{NaBr\left(dư\right)}=0,1\left(mol\right)\\ n_{AgCl}=n_{NaCl}=0,3\left(mol\right)\\ m_{kết.tủa}=m_{AgBr}+m_{AgCl}=188.0,1+0,3.143,5=61,85\left(g\right)\)
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