In the following reaction, how many grams of NaBr will react with 311 grams of Pb(NO3)2?
1 answer:
Molar mass :
NaBr = 103 g/mol
Pb(NO3)2 = 331.20 g/mol
<span><span /><span>Balanced chemical equation :
</span></span>2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr<span>2
</span><span>
2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
g NaBr -------------------> 311 g Pb(NO3)2
331.20 g = 2*103*311
331.20 g = 64066
mass ( NaBr ) = 64066 / 331.20
mass ( naBr) = 193,43 g of NaBr
hope this helps!.
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