Question

In the following reaction, how many grams of NaBr will produce 244 grams of NaNO3? Pb(NO3)20) + 2 NaBr jaPbBr2s 2 NaNO3a) The molar mass of NaBr is 102 9 grams and that of NäNO is 85 grams

Answer 1

 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