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3 years ago

How many grams of lead (ll) sulfate would be produced from the complete reaction of 23.6 g lead (iV) oxide ?

Chemistry

1 answer:

djverab [1.8K]3 years ago

6 0

Answer:

29.92grams of PbSO4

Explanation:

lead (iV) oxide = PbO2 = Molar mass: 239.2 g/mol

lead (ll) sulfate = PbSO4 = Molar mass: 303.26 g/mol

PbO2 = PbSO4

1:1 ratio

Pb = Lead

Lead has an oxidation number of 4+

O = Oxygen

Oxygen has an oxidation number of 2-

PbO2 + 4H+ + SO4 2- + 2e- = PbSO4(s) + 2H2O

Ok so the above would be the likely complete reaction, though we don't really need this as we already know the ratio is 1:1.

23.6g of PbO2

23.6/239.2 = 0.09866 Moles of PbO2

Since we have a 1:1 ratio we know that the same number of moles of PbSO4 are produced and since we know the molar mass it's simply molar mass multiplied by number of moles.

303.26 x 0.09866 = 29.92grams of PbSO4

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Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$

(2) $XCl_4(s)\rightarrow X(s)+2Cl_2(g)$ $\Delta H_2=-461.9kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$ $\Delta H_3=4\times -92.3kJ=-369.2kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $2H_2O(l)\rightarrow 2H_2O(g)$ $\Delta H_5=2\times 44.0kJ=88.0kJ$

The expression for enthalpy of main reaction will be:

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5$

$\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)$

$\Delta H=-1048.6kJ$

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

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