!00.0 grams of Na2SO4 reacts with 150.00 grams of Ba(NO3)2to form NaNO3 and BaSO4. How many grams of barium sulfate will form?
please help i really need the help
This is a limiting reagent problem. I know because amounts are given for BOTH reactants.
mols Na2SO4 = grams/molar mass.
mols Ba(NO3)2 = grams/molar mass.
Convert mols Na2SO4 to mols of the product. Do the same for mols Ba(NO3)2 to the product. It is likely that these two numbers will not agree which means one of them must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value. Use the smaller valuer and convert to grams of the product.
g BaSO4 = mols BaSO4 x molar mass BaSO4.
To determine the amount of barium sulfate (BaSO4) that will form, we need to use stoichiometry and balanced chemical equations.
The balanced chemical equation for the reaction between Na2SO4 and Ba(NO3)2 is:
Na2SO4 + Ba(NO3)2 -> NaNO3 + BaSO4
From the equation, we can see that the stoichiometric ratio of Na2SO4 to BaSO4 is 1:1. This means that for every 1 mole of Na2SO4, 1 mole of BaSO4 will be formed.
Step 1: Convert the given masses to moles.
- Moles of Na2SO4 = mass of Na2SO4 / molar mass of Na2SO4
- Moles of Ba(NO3)2 = mass of Ba(NO3)2 / molar mass of Ba(NO3)2
Step 2: Determine the limiting reactant.
The limiting reactant is the one that is completely consumed and determines the amount of product formed. We need to compare the moles of the reactants and identify the one that forms the fewest moles of the product. The reactant that forms the fewest moles of BaSO4 will be the limiting reactant.
Step 3: Calculate the moles of BaSO4 formed.
Since the stoichiometric ratio between Na2SO4 and BaSO4 is 1:1, the number of moles of BaSO4 formed will be equal to the number of moles of the limiting reactant.
Step 4: Convert moles of BaSO4 to grams.
- Mass of BaSO4 = moles of BaSO4 x molar mass of BaSO4
Let's calculate each step:
Step 1:
- Moles of Na2SO4 = 100.0 g / (22.99 g/mol + 32.06 g/mol + 4 * 16.00 g/mol) = 0.4429 mol
- Moles of Ba(NO3)2 = 150.00 g / (137.33 g/mol + 2 * 14.01 g/mol + 6 * 16.00 g/mol) = 0.7648 mol
Step 2:
To determine the limiting reactant, we compare the mole ratio of the reactants based on the balanced equation. In this case, the mole ratio is 1:1 for Na2SO4 and Ba(NO3)2. Since the moles of Na2SO4 is less than the moles of Ba(NO3)2, Na2SO4 is the limiting reactant.
Step 3:
The moles of BaSO4 formed will be equal to the moles of Na2SO4, which is 0.4429 mol.
Step 4:
- Mass of BaSO4 = 0.4429 mol x (137.33 g/mol + 32.06 g/mol + 4 * 16.00 g/mol) = 147.57 g
Therefore, 147.57 grams of barium sulfate (BaSO4) will form.