What mass of Cu metal is required to react with 6.8 drops of 16M HNO3? Assume 20 drops = 1 mL. Cu (MM 63.54 g/mol), (16M = 16 mol/L) Cu(s) + 4HNO3(aq) Cu(NO3)2(aq) + 2NO2(g) + 2H2O(1) Hint: x drops HNO, X. Cu mL X LX. mol X. mol X. drops mL L mol mol Give answer to 2 significant figures. Do not use scientific notation. After the number, include one space and then the unit, g. Do calculation all at once, round off at the end.