# In the following reaction, how many grams of lead(II) nitrate Pb(NO3)2 will produce 425 grams of sodium nitrate (NaNO3)?

Pb(NO3)2(aq) + 2 NaBr(aq) PbBr2(s) + 2 NaNO3(aq)

The molar mass of Pb(NO3)2 is 331.21 grams and that of NaNO3 is 85 grams.

## mols NaNO3 needed = grams/molar mass.

Using the coefficients in the balanced equation, convert mols NaNO3 to mols Pb(NO3)2. In this case that's mols NaNO3 x (1 mol Pb(NO3)2/2 mols NaNO3) = x mols NaNO3 x (1/2) = ?

## To determine the number of grams of lead(II) nitrate (Pb(NO3)2) needed to produce 425 grams of sodium nitrate (NaNO3), follow these steps:

1. Calculate the molar mass of NaNO3, which is 85 grams/mol.

2. Using the balanced chemical equation, 1 mole of Pb(NO3)2 reacts to produce 2 moles of NaNO3.

3. Divide the given mass of NaNO3 (425 grams) by the molar mass of NaNO3 to convert it to moles: 425 g / 85 g/mol = 5 moles of NaNO3.

4. Since 1 mole of Pb(NO3)2 produces 2 moles of NaNO3, you'll need half as many moles of Pb(NO3)2: 5 moles of NaNO3 / 2 = 2.5 moles of Pb(NO3)2.

5. Finally, multiply the moles of Pb(NO3)2 by its molar mass (331.21 g/mol) to find the mass needed: 2.5 moles * 331.21 g/mol = 828.03 grams.

Therefore, you will need 828.03 grams of lead(II) nitrate (Pb(NO3)2) to produce 425 grams of sodium nitrate (NaNO3).

## To find the number of grams of lead(II) nitrate (Pb(NO3)2) that will produce 425 grams of sodium nitrate (NaNO3), we need to use stoichiometry.

In the balanced chemical equation provided:

Pb(NO3)2(aq) + 2 NaBr(aq) -> PbBr2(s) + 2 NaNO3(aq)

We can see that the ratio of Pb(NO3)2 to NaNO3 is 1:2. This means that for every 1 mole of Pb(NO3)2, 2 moles of NaNO3 are produced.

Step 1: Convert the given mass of NaNO3 to moles.

Moles of NaNO3 = Mass of NaNO3 / Molar mass of NaNO3

Molar mass of NaNO3 = 85 grams/mol

Moles of NaNO3 = 425 grams / 85 grams/mol

Moles of NaNO3 = 5 moles

Step 2: Use the ratio of stoichiometry to find the moles of Pb(NO3)2.

Moles of Pb(NO3)2 = (Moles of NaNO3) / (moles of NaNO3 produced per mole of Pb(NO3)2)

From the balanced equation, we know that 1 mole of Pb(NO3)2 produces 2 moles of NaNO3.

Moles of Pb(NO3)2 = 5 moles / 2

Moles of Pb(NO3)2 = 2.5 moles

Step 3: Convert moles of Pb(NO3)2 to grams.

Mass of Pb(NO3)2 = (Moles of Pb(NO3)2) x (Molar mass of Pb(NO3)2)

Molar mass of Pb(NO3)2 = 331.21 grams/mol

Mass of Pb(NO3)2 = 2.5 moles x 331.21 grams/mol

Mass of Pb(NO3)2 = 828.03 grams

Therefore, 425 grams of sodium nitrate (NaNO3) will produce approximately 828.03 grams of lead(II) nitrate (Pb(NO3)2).