If 5.14 mol of Cu(NO3)2 are reacted with an excess of NaCl, how many grams of copper (II) chloride should be produced? (Round to the tenths place)

Cu(NO3)2 + 2 NaCl -> CuCl2 + 2 NaNO3

True, the standard unit for volume is liters which can be represented by either a lower case "l" or a capital "l"

H tricky one let me think and get back to u

Using dilution equationm1v1 =m2v2 ; where m1 and m1 are the intiial and final concentration, and v1 and v2 are the initial and final volume.5  × 5 m = 8 l  × m2m2 = 25/8 = 3.125 mtherefore; the resulting molarity is 3.125 m