a 1.009g mixture of the solid salts Na2SO4 and forms an aqueous solution with the precipitation of PbSO4. The precipitate was filtered and dried and its mass was determined to be .471 g. the limiting reactant was determined to be Na2SO4. Write the molecular form of the equation for the reaction?

Write the net ionic equation

how many moles and grams of na2so4 are in the reaction mixture.

how many moles and grams of pb(no3)2 reacted in the reaction mixture?

what is the percent by mass of each salt in the mixture?

The molecular equation is

Pb(NO3)2(aq) + Na2SO4(aq) ==> PbSO4(s) + 2NaNO3(aq)

net ionic equation is
Pb^+2(aq) + SO4^-2(aq) ==> PbSO4(s)

You know you have 0.471 g PbSO4. Convert that to moles PbSO4, then convert to moles Pb(NO3)2 (since that was the limiting reagent). Convert moles Pb(NO3)2 to grams Pb(NO3)2 which you must have had initially. Then
mass percent Pb(NO3)3 = (grams Pb(NO3)2/mass sample)*100 = ??
1.009 g sample - g Pb(NO3)2 = grams Na2SO4 and convert that to mass percent the same as above.

To answer these questions, let's break it down step by step:

1. Write the molecular form of the equation for the reaction:
The given information tells us that Na2SO4 reacts with Pb(NO3)2 to form PbSO4 as a precipitate. The molecular equation can be written as:
2 Na2SO4 + Pb(NO3)2 → 2 NaNO3 + PbSO4

2. Write the net ionic equation:
To write the net ionic equation, we focus only on the species directly involved in the reaction. In this case, we can leave out the spectator ions (ions that remain unchanged). The net ionic equation is:
2 Na+ + SO4²⁻ + Pb²⁺ → PbSO4

3. Calculate the moles and grams of Na2SO4 in the reaction mixture:
Given that the total mass of the mixture is 1.009 g and Na2SO4 is the limiting reactant, we can calculate the moles and grams of Na2SO4 using its molar mass.

Molar mass of Na2SO4 = 22.990 g/mol (2 Na) + 32.060 g/mol (1 S) + (4 × 16.00 g/mol) (4 O)
= 142.044 g/mol

Moles of Na2SO4 = Mass / Molar mass
= 1.009 g / 142.044 g/mol
≈ 0.0071 mol

Grams of Na2SO4 = Moles × Molar mass
= 0.0071 mol × 142.044 g/mol
≈ 1.0115 g

So, there are approximately 0.0071 mol or 1.0115 g of Na2SO4 in the reaction mixture.

4. Calculate the moles and grams of Pb(NO3)2 reacted in the reaction mixture:
Given that Na2SO4 is the limiting reactant, we can use the stoichiometry of the reaction to determine the moles and grams of Pb(NO3)2 reacted.

From the balanced equation, the ratio of Na2SO4 to Pb(NO3)2 is 2:1.
Therefore, the moles of Pb(NO3)2 reacted = (moles of Na2SO4) / 2
= 0.0071 mol / 2
= 0.00355 mol

Grams of Pb(NO3)2 reacted = Moles × Molar mass
= 0.00355 mol × (207.20 g/mol + 3 × 16.00 g/mol)
≈ 0.7495 g

So, there are approximately 0.00355 mol or 0.7495 g of Pb(NO3)2 reacted in the reaction mixture.

5. Calculate the percent by mass of each salt in the mixture:
The percent by mass can be calculated using the following formula:
Percent by mass = (mass of solute ÷ mass of mixture) × 100

For Na2SO4:
Percent by mass = (1.0115 g ÷ 1.009 g) × 100
≈ 100.28%

For PbSO4:
Percent by mass = (0.471 g ÷ 1.009 g) × 100
≈ 46.62%

So, in the mixture, Na2SO4 constitutes approximately 100.28% and PbSO4 constitutes approximately 46.62% by mass. Please note that the total percent exceeds 100% due to experimental error.

The molecular form of the equation for the reaction is:

2 Na2SO4 + Pb(NO3)2 -> 2 NaNO3 + PbSO4

The net ionic equation for the reaction is:

2 Na+ (aq) + SO4²- (aq) + Pb²+ (aq) + 2 NO3- (aq) -> 2 Na+ (aq) + 2 NO3- (aq) + PbSO4 (s)

To find the number of moles and grams of Na2SO4 in the reaction mixture, we can use the molar mass of Na2SO4.
The molar mass of Na2SO4 is 22.99 g/mol (Na) + 32.07 g/mol (S) + (4 x 16.00 g/mol) (O) = 142.04 g/mol.

To find the number of moles of Na2SO4, we divide the mass of the mixture (1.009 g) by the molar mass of Na2SO4:

moles of Na2SO4 = 1.009 g / 142.04 g/mol = 0.00711 mol

To find the grams of Na2SO4, we multiply the number of moles by the molar mass:

grams of Na2SO4 = 0.00711 mol x 142.04 g/mol = 1.01 g

To find the number of moles and grams of Pb(NO3)2 reacted in the mixture, we can use the molar mass of Pb(NO3)2.

The molar mass of Pb(NO3)2 is 207.2 g/mol (Pb) + (2 x 14.01 g/mol) (N) + (6 x 16.00 g/mol) (O) = 331.21 g/mol.

Since Na2SO4 is the limiting reactant, the molar ratio between Na2SO4 and Pb(NO3)2 is 2:1. Therefore:

moles of Pb(NO3)2 = 0.00711 mol / 2 = 0.00356 mol

grams of Pb(NO3)2 = 0.00356 mol x 331.21 g/mol = 1.18 g

To calculate the percent by mass of each salt in the mixture, we need the total mass of the mixture, which is the sum of the masses of Na2SO4 and Pb(NO3)2:

total mass of mixture = mass of Na2SO4 + mass of Pb(NO3)2 = 1.009 g + 1.18 g = 2.189 g

Percent by mass of Na2SO4 = (mass of Na2SO4 / total mass of mixture) x 100%
= (1.009 g / 2.189 g) x 100% = 46.1%

Percent by mass of Pb(NO3)2 = (mass of Pb(NO3)2 / total mass of mixture) x 100%
= (1.18 g / 2.189 g) x 100% = 53.9%