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How many grams of lead(II) nitrate, pb(NO3)2,are present in 4.05 moles

1 Answer

7 votes

Answer:

1341.1737 gm in 4.05 moles

Step-by-step explanation:

From periodic table

mole weight pb = 207.2 gm

N 14.007 *2 = 28.014 g,

O 15.99 *6 = 95.94 gm

total= 331.154 gm per mole * 4.05 moles =

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