Question

Answer 1

 

\(N_{Fe}=0,25\cdot6\cdot10^{23}=1,5\cdot10^{23}\)

\(N_{Al}=1,5\cdot6\cdot10^{23}=9\cdot10^{23}\\ N_{H_2}=0,5\cdot6\cdot10^{23}=3\cdot10^{23}\\ N_{NaCl}=0,25\cdot6\cdot10^{23}=1,5\cdot10^{23}\\ n_{H_2O}=\dfrac{1,2\cdot10^{22}}{6\cdot10^{23}}=0,02\left(mol\right)\\ n_{Mg}=\dfrac{0,72\cdot10^{23}}{6\cdot10^{23}}=0,12\left(mol\right)\\ n_{CO_2}=\dfrac{3\cdot10^{22}}{6\cdot10^{23}}=0,05\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{80}{56\cdot2+16\cdot3}=0,5\left(mol\right)\\ n_{CaCO_3}=\dfrac{10}{40+12+16\cdot3}=0,1\left(mol\right)\\ n_{Na_2SO_4}=\dfrac{56,8}{23\cdot2+32+16\cdot4}=0,4\left(mol\right)\\ n_{K_2SO_4}=\dfrac{43,5}{39\cdot2+32+16\cdot4}=0,25\left(mol\right)\\ n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\\ n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)

\(m_{CuO}=0,25\cdot\left(64+16\right)=20\left(g\right)\\ m_{BaCl_2}=0,25\cdot\left(137+35,5\cdot2\right)=52\left(g\right)\\ m_{NaCl}=0,05\cdot\left(23+35,5\right)=2,925\left(g\right)\\ n_{H_2SO_4}=0,15\cdot\left(2+32+16\cdot4\right)=14,7\left(g\right)\)