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).