Convert the following word equation into a balanced chemical equation:aluminum metal copper(II) fluoride → aluminum fluoride copper metalThe answer is: 2Al(s) 3CuF2(aq) → 2AlF3(aq) 3Cu(s) ( I'm confused on how they got the answer plz explain)
Here, aluminum metal and copper metals are mixed but aluminum has higher solubility than copper isYour text book will have that solubility chartWhen Al is contact with F ions, F ions will combine with Al and form AlF leaving Cu by itself and expressed as solidIn terms of balancing equation, you must count each elements and find out the right coefficient to equal out the number of each element from left side to right side(reactant to product) Start from thisAl + CuF2 - AlF3 + Cu As you see you can see there are 2 F on reactant side and 3 F on product sideThe least common multiple of 2 and 3 are 6In order for us to have 6 F on each side, we need to multiply 3 on CuF2 and 2 on AlF3That makesAl + 3 CuF2 - 2 AlF3 + Cu Now number of F is same on both sideBut because of the added coefficient, reactant side Cu is now 3, so you add coefficient of 3 to product side CuAl + 3 CuF2 - 2 AlF3 + 3 Cu Now, AlF3 has coefficient of 2 which makes 2 Al on product sideSo, for us to balance that out, we add coefficient of 2 on reactant side of Al2 Al + 3 CuF2 - 2 AlF3 + 3 Cu H.
