How many grams of lead(II) sulfate are formed when 150.0 g of lead reacts?

Question

asked Jan 9, 2021 40.0k views

1 Answer

Narendra Sorathiya · Answer 1 · 2021-01-16T02:02:24+0000

Answer:

219.75g

Step-by-step 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

⇒
(150)/(207) = 0.7246 moles

∴ mass of PbSO4 produced =

303.26 × 0.7246 = 219.75g