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3 years ago

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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.

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:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

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$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

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$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

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The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

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First of all, you have to translate the words into an equation.

Fe(iii)2O3 + C ==> Fe + CO2

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2 Fe(iii)2O3 + C ==> 2 Fe + 3 CO2

Finally it is the turn of the carbons. There are 3 on the right, so you must make the carbon on the left = 3

2 Fe(iii)2O3 + 3 C ==> 2 Fe + 3 CO2

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Answer:

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Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

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Suppose 31.57% = 31.57 grams

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moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

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moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

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Step 5: We divide by the smallest number of moles

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H: 5.25/2.63 = 2 → 4

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The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

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