Find the mass of AlCl3 that isproducednwhen 25.0 g Al03 REACT WITH EXCESS HCL AC

Question

Find the mass of AlCl3 that isproducednwhen 25.0 g Al03 REACT WITH EXCESS HCL ACCORDING TO THEBELOW EQUATION BALANCED EQUATION Al2O3 + 6CHL ----->   2AlCL3 + 3H20 A. 155   B. 72.9 C.  125 D. 32.6    E. 16.3 Find the mass of AlCl3 that isproducednwhen 25.0 g Al03 REACT WITH EXCESS HCL ACCORDING TO THEBELOW EQUATION BALANCED EQUATION Al2O3 + 6CHL ----->   2AlCL3 + 3H20 A. 155   B. 72.9 C.  125 D. 32.6    E. 16.3 Al2O3 + 6CHL ----->   2AlCL3 + 3H20 A. 155   B. 72.9 C.  125 D. 32.6    E. 16.3

Explanation / Answer

Al2O3 + 6HCl ----->   2AlCl3 + 3 H20
Mol. wt. of Al2O3 is = 2(27) + 3 ( 16 ) = 102 g Mol. wt. of AlCl3 is = 27 + 3 ( 35.5 ) = 133.5 g According to the equation 102 g of Al2O3 reacts with HClproduces 2 * 133.5 g of AlCl3                                         25g of Al2O3 reacts with HCl produces X g of AlCl3                                                                 X = ( 2* 133.5 * 25 ) / 102                                                                     = 65.4411 g

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