\(a.\\ n_{CaO}=a;n_{CaCO_3}=b\\ BTNT:\\ a+b=\dfrac{33,3}{111}=0,3\\ b=\dfrac{4,958}{24,79}=0,2\\ a=0,1\\ m_{CaO}=56a=5,6g\\ m_{CaCO_3}=100b=20g\\ b.C_M=\dfrac{0,6}{0,3}=2mol\cdot L^{-1}\)
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\(n_{H_2}=\dfrac{8,6765}{24,79}=0,35mol\\ m_{muối}=10,14-1,54+0,7\cdot35,5=33,45g=m\)
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