Question

A 1.500 g dried sample contains only NaI and NaBr. It is precipitated with excess AgNO3. The dried ppt weighs 2.4782 g. What is the % NaI in the original sample? (note AgBr and AgI are the precipitates) AgBr = 187.772; AgI = 234.772; NaI = 149.904; NaBr = 102.904

Answer 1

m = 1.5 g sample

NaI + NaBr

m = 2.4782 g of precipitates

mol of NaI * MW + mol of NaBr * MW = 1.5

x* 149.904+ y * 102.904= 1.5 (EQUATION 1)

for equation 2:

mol of AgI * MW + mol of AgBr* MW = 2.4782

x * 234.772 + y * 187.77 = 2.4782 (EQUATION 2)

2 equations = 2 unkown, this can be solved

x* 149.904+ y * 102.904= 1.5

y = (1.5 - x* 149.904) / (102.904)

y = 1.5/102.904 - 149.904/102.904x

y = 0.014576 -1.45673x

substitute in 2

y = 0.014576 -1.45673x

x * 234.772 + ( 0.014576 -1.45673x) * 187.77 = 2.4782

234.772x + 187.77 *0.014576 - 187.77 *1.45673x = 2.4782

x ( 234.772 - 273.53) = 2.4782-187.77 *0.014576

x = (-0.25873) / -38.75

x = 0.0066769 mol of NaI

y = 0.014576 -1.45673x

y = 0.014576 -1.45673*0.0066769

y = 0.00484955mol of NaBr