how many moles of lead (II) sulfate will be produced when 500 grams of magnesium sulfate is allowedto react with lead (II) nitrate?
A worked example. Just follow the steps.
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To determine the number of moles of lead (II) sulfate produced, we need to calculate the molar ratio between magnesium sulfate (MgSO4) and lead (II) sulfate (PbSO4) using their balanced chemical equation. The balanced equation for the reaction between magnesium sulfate and lead (II) nitrate is as follows:
MgSO4 + Pb(NO3)2 → PbSO4 + Mg(NO3)2
From the balanced equation, we can see that the molar ratio between magnesium sulfate and lead (II) sulfate is 1:1. This means that for every 1 mole of magnesium sulfate reacted, we will produce 1 mole of lead (II) sulfate.
To find the number of moles of magnesium sulfate, we first need to calculate its molar mass. The molar mass of magnesium sulfate is calculated by adding up the atomic masses of its constituent elements:
Mg: 24.31 g/mol
S: 32.06 g/mol
O: 16.00 g/mol (x 4 since there are four oxygen atoms in MgSO4)
Molar mass of MgSO4 = 24.31 + 32.06 + (16.00 x 4) = 120.37 g/mol
Now, we can calculate the number of moles of magnesium sulfate by dividing its mass by its molar mass:
Moles of MgSO4 = mass of MgSO4 / molar mass of MgSO4
= 500 g / 120.37 g/mol
Calculating this value, we find that there are approximately 4.15 moles of magnesium sulfate.
Since the molar ratio between magnesium sulfate and lead (II) sulfate is 1:1, the number of moles of lead (II) sulfate produced will also be 4.15 moles.