What mass of lead (II) sulphate would be produced by the action of excess dilute sulpuric acid on 10 g of lead nitrate dissolved in water? I need help, at least an example.
To determine the mass of lead (II) sulphate produced, we need to calculate the stoichiometry of the reaction between lead nitrate (Pb(NO3)2) and sulphuric acid (H2SO4). The balanced chemical equation for the reaction is as follows:
Pb(NO3)2 + H2SO4 → PbSO4 + 2 HNO3
From the equation, we can see that one mole of lead nitrate reacts with one mole of sulphuric acid to produce one mole of lead (II) sulphate. This means the mole ratio between lead nitrate and lead (II) sulphate is 1:1.
To solve this problem, we can follow these steps:
Step 1: Calculate the number of moles of lead nitrate (Pb(NO3)2).
Moles = mass / molar mass
Molar mass of Pb(NO3)2:
Lead (Pb) = 207.2 g/mol
Nitrate (NO3) = (14 + 3 * 16) g/mol = 62 g/mol
Molar mass of Pb(NO3)2 = (207.2 + 2 * 62) g/mol = 331.2 g/mol
Number of moles of Pb(NO3)2 = 10 g / 331.2 g/mol
Step 2: Since the reaction is given as having excess sulphuric acid, we know that all the moles of lead nitrate react to produce the same number of moles of lead (II) sulphate.
Number of moles of PbSO4 = Number of moles of Pb(NO3)2 = moles calculated in Step 1.
Step 3: Calculate the mass of lead (II) sulphate (PbSO4).
Mass = moles of PbSO4 * molar mass of PbSO4
Molar mass of PbSO4:
Lead (Pb) = 207.2 g/mol
Sulphate (SO4) = (32 + 4 * 16) g/mol = 96 g/mol
Molar mass of PbSO4 = (207.2 + 96) g/mol = 303.2 g/mol
Mass of PbSO4 = moles of PbSO4 * 303.2 g/mol (moles calculated in Step 2)
By following these steps and performing the calculations, you can determine the mass of lead (II) sulphate that would be produced when 10 g of lead nitrate reacts with excess sulphuric acid.
Note: The molar masses provided are for illustrative purposes only, and you should use the most accurate values available in your calculations.