For the reaction Pb (NO3) 2 + 2KI → PbI2 + 2KNO3, how many moles of lead iodide are produced from 246.2 g of potassium iodide?

Question

Chemistry

Elaine Terrell

20 February, 16:26

For the reaction Pb (NO3) 2 + 2KI → PbI2 + 2KNO3, how many moles of lead iodide are produced from 246.2 g of potassium iodide?

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Answers (1)

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Moshe Duffy · Answer 1

M (KI) = 246.2 g

M (KI) = 166.0 g/mol

n (KI) = m (KI) / M (KI)

n (KI) = 246.2/166.0≈1.48 mol

2 mol KI - 1 mol PbI₂

1.48 mol KI - x mol PbI₂

x=1.48*1/2=0.74 mol