Lead(II) nitrate reacts with potassium iodine to produce lead(II) iodide and Potassium nitrate. If 1.23 mg of lead nitrate are consumed, what is the mass of the potassium nitrate produced?

Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3

Convert 1.23 mg Pb(NO3)2 to moles. moles = grams/molar mass

Using the coefficients in the balanced equation, convert moles Pb(NO3)2 to moles KNO3.

Now convert moles KNO3 to grams. g = moles x molar mass.

It is reversible I am a high school teacher

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To determine the mass of potassium nitrate produced when 1.23 mg of lead(II) nitrate reacts, we need to first understand the stoichiometry of the reaction.

The balanced equation for the reaction is as follows:

Pb(NO3)2 + 2 KI -> PbI2 + 2 KNO3

From the equation, we can see that each mole of lead(II) nitrate reacts with 2 moles of potassium iodide to produce 1 mole of lead(II) iodide and 2 moles of potassium nitrate.

1. Convert the mass of lead nitrate consumed to moles using its molar mass.
The molar mass of Pb(NO3)2 can be calculated as follows:
Pb: atomic mass = 207.2 g/mol
N: atomic mass = 14.0 g/mol
O: atomic mass = 16.0 g/mol
Total molar mass of Pb(NO3)2 = (207.2 g/mol) + 2 * (14.0 g/mol + 3 * 16.0 g/mol) = 331.2 g/mol

Now, calculate the number of moles of Pb(NO3)2:
moles = mass / molar mass = 1.23 mg / 331.2 g/mol

2. Determine the moles of potassium nitrate produced.
Since the stoichiometry of the balanced equation indicates that 1 mole of lead(II) nitrate produces 2 moles of potassium nitrate, the moles of potassium nitrate will be twice the moles of lead(II) nitrate.

moles of KNO3 produced = 2 * moles of Pb(NO3)2

3. Convert the moles of potassium nitrate produced to mass using its molar mass.
The molar mass of KNO3 can be calculated as follows:
K: atomic mass = 39.1 g/mol
N: atomic mass = 14.0 g/mol
O: atomic mass = 16.0 g/mol
Total molar mass of KNO3 = 39.1 g/mol + 14.0 g/mol + 3 * 16.0 g/mol = 101.1 g/mol

Now, calculate the mass of KNO3 produced:
mass = moles * molar mass

By following these steps, you can calculate the mass of potassium nitrate produced when 1.23 mg of lead nitrate reacts.