smithmariah73901 smithmariah73901
  • 13-05-2020
  • Chemistry
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How many grams of lead(II) sulfate are formed when 150.0 g of lead reacts?

Respuesta :

jamesesther1234
jamesesther1234 jamesesther1234
  • 16-05-2020

Answer:

219.75g

Explanation:

The balanced equation for this reaction is:

Pb + H2SO4 → PbSO4 + H2

mole ratio of lead to lead (II) sulfate is 1 : 1

1 mole of Pb = 207g

?? moles of Pb - 150g

⇒[tex]\frac{150}{207}[/tex] = 0.7246 moles

∴ mass of PbSO4 produced =

303.26 × 0.7246 = 219.75g

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