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Oxana [17]
3 years ago
5

In the following reaction, how many grams of NaBr will react with 311 grams of Pb(NO3)2?

Chemistry
1 answer:
Mashutka [201]3 years ago
5 0
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!.


</span>
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