Question

If 100.0ml of 0.495 m na2so4 are added to 100.0ml of 0.773 m pb(no3)2, how many grams of pbso4 can be produced

Answer 1

To determine how many grams of PbSO4 can be produced, identify the limiting reagent (in this case, Na2SO4) and then use stoichiometry to convert moles of the limiting reagent to grams of product. The maximum amount of PbSO4 that can be produced from 100.0 ml of 0.495 M Na2SO4 and 100.0 ml of 0.773 M Pb(NO3)2 is 15.01335 grams.

Step-by-step explanation:

The question is asking how many grams of lead sulfate (PbSO4) can be produced when solutions of sodium sulfate (Na2SO4) and lead nitrate (Pb(NO3)2) are mixed. This is a stoichiometry problem involving a precipitation reaction, which is a common topic in high school chemistry courses.

To answer this question, we need to follow these steps:

1. Write the balanced chemical equation for the reaction.
2. Calculate the moles of Na2SO4 and Pb(NO3)2 that are present.
3. Determine the limiting reagent.
4. Use the molar ratio from the balanced equation to find the number of moles of PbSO4 that can be produced.
5. Convert the moles of PbSO4 to grams.

Here's the balanced equation for the reaction:

Na2SO4(aq) + Pb(NO3)2(aq) → PbSO4(s) + 2NaNO3(aq)

Next, calculate the moles of each reactant:

• 0.495 M Na2SO4 × 0.100 L = 0.0495 moles Na2SO4
• 0.773 M Pb(NO3)2 × 0.100 L = 0.0773 moles Pb(NO3)2

Since Na2SO4 reacts with Pb(NO3)2 in a 1:1 mole ratio and there are fewer moles of Na2SO4 than Pb(NO3)2, Na2SO4 is the limiting reagent. Therefore, the reaction will produce 0.0495 moles of PbSO4.

The molar mass of PbSO4 is 303.3 g/mol, so the mass of PbSO4 that can be produced is:

0.0495 moles × 303.3 g/mol = 15.01335 grams of PbSO4

Answer 2

To determine how many grams of PbSO4 can be produced, identify the limiting reagent (in this case, Na2SO4) and then use stoichiometry to convert moles of the limiting reagent to grams of product. The maximum amount of PbSO4 that can be produced from 100.0 ml of 0.495 M Na2SO4 and 100.0 ml of 0.773 M Pb(NO3)2 is 15.01335 grams.

Step-by-step explanation:

The question is asking how many grams of lead sulfate (PbSO4) can be produced when solutions of sodium sulfate (Na2SO4) and lead nitrate (Pb(NO3)2) are mixed. This is a stoichiometry problem involving a precipitation reaction, which is a common topic in high school chemistry courses.

To answer this question, we need to follow these steps:

1. Write the balanced chemical equation for the reaction.
2. Calculate the moles of Na2SO4 and Pb(NO3)2 that are present.
3. Determine the limiting reagent.
4. Use the molar ratio from the balanced equation to find the number of moles of PbSO4 that can be produced.
5. Convert the moles of PbSO4 to grams.

Here's the balanced equation for the reaction:

Na2SO4(aq) + Pb(NO3)2(aq) → PbSO4(s) + 2NaNO3(aq)

Next, calculate the moles of each reactant:

• 0.495 M Na2SO4 × 0.100 L = 0.0495 moles Na2SO4
• 0.773 M Pb(NO3)2 × 0.100 L = 0.0773 moles Pb(NO3)2

Since Na2SO4 reacts with Pb(NO3)2 in a 1:1 mole ratio and there are fewer moles of Na2SO4 than Pb(NO3)2, Na2SO4 is the limiting reagent. Therefore, the reaction will produce 0.0495 moles of PbSO4.

The molar mass of PbSO4 is 303.3 g/mol, so the mass of PbSO4 that can be produced is:

0.0495 moles × 303.3 g/mol = 15.01335 grams of PbSO4