Please solve second one

Dear Student, Na2SO4 + 2AgNO3 → Ag2SO4 + 2NaNO31 mol 2 mol 1 mol1 mol of Na2SO4 reacts with 2 mol of AgNO3 to give 1 mol of Ag2SO4Number of moles of Na2SO4 in solution=Molarity×Volume = 0.2×0.5 = 0.1 molNumber of moles of AgNO3 in solution = 0.3×0.25 = 0.075 molNow, 1 mol of Na2SO4 gives 1 mol of Ag2SO4So, 0.1 mol of Na2SO4 gives 0.1 mol of Ag2SO4Similarly, 2 mol of AgNO3 gives 1 mol of Ag2SO4So, 0.075 mol of AgNO3 will give 12×0.075=0.0375 mol of Ag2SO4Since, AgNO3 gives less mol of Ag2SO4, so the limiting reagent here is AgNO3and moles of Ag2SO4 precipitated= 0.0375 mol