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

 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
              

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