All categories

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

You might be interested in

If a substance conducts heat easily, it is considered to be a

GrogVix [38]

It is considered to be a conductor

5 0

3 years ago

Read 2 more answers

I need the answer fast plzzz!! NO LINKS

Arturiano [62]

Answer:

your answer is 12 hope it's correct answer

4 0

3 years ago

B) Which term is defined as a measure of the average kinetic energy of the particles in a sample of matter?

kogti [31]

Entropy is also defined as a measure of the average kinetic energy of particles in a sample of matter.

6 0

3 years ago

Read 2 more answers

A sample of hydrogen occupies a volume of 351 mL at a temperature of 20 degrees Celsius. What is the new volume if the temperatu

xeze [42]

Volume of Hydrogen V1 = 351mL
Temperature T1 = 20 = 20 + 273 = 293 K
Temperature T2 = 38 = 38 + 273 = 311 K
We have V1 x T2 = V2 x T1
So V2 = (V1 x T2) / T1 = (351 x 311) / 293 = 372.56
Volume at 38 C = 373 ml

6 0

3 years ago

0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium

Free_Kalibri [48]

Moles of potassium permanganate = 0.0008

<h3>Further explanation </h3>

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range

Reaction

5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) + 10CO2(g) + 8H2O(1)

The end point ⇒titrant and analyte moles equal

titrant : potassium permanganate-KMnO4

analyte : sodium oxalate - Na2C2O4

so moles of KMnO4 = moles of Na2C2O4

moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{0.264}{134 g/mol}\\\\mol=0.002$

From equation, mol ratio Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :

$\tt \dfrac{2}{5}\times 0.002=0.0008$

6 0

3 years ago