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# Answer to Question #199141 in Chemistry for Jessica Phan

Question #199141
• Calculate the number of moles of lead (II) nitrate
• Calculate the number of moles of sodium iodide
• Calculate the number of moles of lead (II) iodide produced assuming sodium iodide is limiting and lead (II) nitrate is in excess using mole ratios between lead (II) iodide and sodium iodide.
• Calculate the number of moles of lead (II) iodide produced assuming lead (II) nitrate is limiting and sodium iodide is in excess using mole ratios between lead (II) iodide and lead (II) nitrate
• Identify which reactant is the limiting reagent based upon your calculations.
• Calculate the mass of lead (II) iodide

Balanced Equation: Pb(NO3)2 + 2 NaI = Pbl2 + 2 NaNO3

2.190 g of sodium iodide

2.282 g of lead (II) nitrate

The average amount of Lead (II) iodide: 2.9955

1
2021-05-31T02:08:04-0400

Pb(NO3)2 + 2NaI = Pbl2 + 2NaNO3

The number of moles of lead nitrate:

n(Pb(NO3)2) = m / Mr = 2.282 g / 331.2 g/mol = 0.0069 mol

The number of moles of sodium iodide:

n(NaI) = m / Mr = 2.190 g / 149.89 g/mol = 0.0146 mol

The number of moles of lead iodide (sodium iodide is limiting):

n(Pbl2) = n(NaI) / 2 = 0.0146 mol / 2 = 0.0073 mol

The number of moles of lead iodide (lead nitrate is limiting):

n(Pbl2) = n(Pb(NO3)2) = 0.0069 mol

Based on the calculations, lead nitrate is a limiting reagent.

The mass of lead iodide:

m(PbI2) = n(PbI2) × Mr(PbI2) = 0.0069 mol × 461.01 g/mol = 3.18 g

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