Pb(NO3)2 + 2 NaBr ------> PbBr2 + 2 NaNO3
2 mol 2 mol
M(NaBr)= 102 g/mol
M(NaNO3) = 85 g/mol
1) 244g NaNO3 *1 mol NaNO3/85 g NaNO3 = 244/85 mol NaNO3
2) Pb(NO3)2 + 2 NaBr ------> PbBr2 + 2 NaNO3
2 mol 2 mol
x mol 244/85 mol
x=(2 mol* 244/85 mol )/2 mol = 244/85 mol NaBr
3) 244/85 mol NaBr*102g NaBr/1 mol = (244*102/85) g NaBr =292.8 g NaBr