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inna [77]
3 years ago
11

Solutions of sulfuric acid and lead(ll) acetate react to form solid lead(ll) sulfate and a solution of acetic acid. If 7.50 g of

sulfuric acid and 7.50 g of lead(ll) acetate are mixed, calculate the number of grams of sulfuric acid, Iead(ll) acetate, lead(ll)sulfate, and acetic acid present in the mixture after the reaction is complete.
Chemistry
1 answer:
finlep [7]3 years ago
5 0

Answer:

sulfuric acid: 5.24 g

lead(II) acetate: 0 g

lead(II) sulfate: 6.99 g

acetic acid: 2.77 g

Explanation:

The molecular formula of the compounds and their molar masses are:

Sulfuric acid: H2SO4, 98 g/mol;

Lead(II) acetate: Pb(Ac)2, 325 g/mol (Ac = C2H3O2);

Lead(II) sulfate: PbSO4, 303 g/mol;

Acetic acid: HAc, 60 g/mol

Thus, the reaction is:

H2SO4 + Pb(Ac)2 → PbSO4 + 2HAc

The stoichiometry of the reaction is 1:1:1:2. Let's first identify which of the reactant is the limiting, the one that will be totally consumed:

1 mol H2SO4 ------- 1 mol Pb(Ac)2

Transforming the relation to mass, multiplying the number of moles by the molar mass, and assuming H2SO4 as the liting reactant:

98 g H2SO4 ------------ 325 g Pb(Ac)2

7.50 g ------------ x

By a simple direct three rule:

98x = 2437.5

x = 24.87 g of Pb(Ac)2

Thus, there are fewer Pb(Ac)2 than is necessary, so it's the limiting.

So, the mass of sulfuric acid that reacts is:

325 g Pb(Ac)2 ---------------- 98 g H2SO4

7.50g ---------------- x

By a simple direct three rule:

325x = 735

x = 2.26 g of H2SO4

So, the left mass is 7.50 - 2.26 = 5.24 g.

The lead(II) acetate reacts completely, thus the final mass is 0.

The stoichiometry with lead (II) sulfate is:

1 mol Pb(Ac)2 ---------- 1 mol PbSO4

Passing by mass:

325 g Pb(Ac)2 --------------- 303 g PbSO4

7.5 g --------------- x

By a simple direct three rule:

325x = 227.25

x = 6.99 g formed of PbSO4

And the stoichiometry with HAc is:

1 mol Pb(Ac)2 ------- 2 moles HAc

Passing by mass

325 g Pb(Ac)2 ----- 120 g HAc

7.50 g ------ x

By a simple direct three rule:

325x = 900

x = 2.77 g of HAc.

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Answer:

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Explanation:

2SO2(g) + O2(g) + 2H2O(ℓ) −→ 2H2SO4(ℓ)

We want to find the mass in grams of SO2 that is needed to react with 1527 g of O2. First we must convert the grams of O2 to moles of O2 then to moles of SO2 and then to grams of SO2

So first lets find the molar mass of O2

The mass of oxygen according to a periodic table is 15.999

Using this the mass of O2 would be 15.999(2) = 31.988g

Next we need to identify the mole ratio of O2 to SO2

Looking at the equation for 1 mole of O2 there are two moles of SO2

Next we need to find the molar mass of SO2

Again the mass of oxygen is 15.999g and the mass of Sulfur is 32.066

So the mass of SO2 would be 15.999(2) + 32.066 = 64.064g

Now that we have found all the needed conversions :

  • 1 mol O2 = 31.988g
  • 1 mol O2 = 2 mol SO2
  • 1 mol SO2 = 64.064g

We can now use dimensional analysis to calculate the answer.

Kindly check the attached image to see the table. ( sorry if its a bit blurry )
Explanation : The conversions are used to cancel out the units to get to the final unit which is gSO2.

Once the units are cancelled out except for the gSO2 we mutliply and divide based off of what the table says to do.
Here first we divide 1527 by 31.988. We than multiply by 2. Finally we multiply by 64.064 to get the final answer which is 6116gSO2

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