Daniel asks…How to find the mass of the precipitate?130.4mL of 0.459mol/L AgNO3 and 85.23mL of 0.251mol/L AlCl3 are mixed.
3 AgNO3(aq) + AlCl3 (aq) -> Al(NO3)3 (aq) + 3AgCl (s)
What mass of AgCl(s) precipitates? Please show steps and thanks!Nancy answers:Find moles:
0.1304L of 0.459mol/L AgNO3 = 0.05985 moles of AgNO3
0.08523mL of 0.251mol/L AlCl3 = 0.02139 moles of AlCl3
by the equation
3 AgNO3(aq) + AlCl3 (aq) -> Al(NO3)3 (aq) + 3AgCl (s)
they needed to add three times as many
