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3 years ago

2Pb(s) + O2(aq) + 4H+(aq) → 2H2O(l) + 2Pb2+(aq)

Chemistry

2 answers:

defon3 years ago

8 0

Answer:

The answer to your question is 0.269 g of Pb

Explanation:

Data

Lead solution = 0.000013 M

Volume = 100 L

mass = 0.269 g

atomic mass Pb = 207.2 g

Chemical reaction

2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)

Process

1.- Calculate the mass of Pb in solution

Formula

Molarity = $\frac{number of moles}{volume}$

Solve for number of moles

Number of moles = Volume x Molarity

Substitution

Number of moles = 100 x 0.000013

Number of moles = 0.0013

2.- Calculate the mass of Pb formed.

207.2 g of Pb ----------------- 1 mol

x g ----------------- 0.0013 moles

x = (0.0013 x 207.2) / 1

x = 0.269 g of Pb

Step2247 [10]3 years ago

4 0

Answer:

0.269 grams of lead metal reacted

Explanation:

Step 1: Data given

Volume of a 0.000013M lead solution = 100 L

Molar mass of Pb = 207.2 g/mol

Step 2: The balanced equation

2Pb(s) + O2(aq) + 4H+(aq) → 2H2O(l) + 2Pb^2+(aq)

Step 3: Calculate moles of lead

Moles Pb^2+ = molarity * volume

Moles Pb^2+ = 0.000013M * 100L

Moles Pb^2+ = 0.00130 moles

Step 4: Calculate moles Pb metal

For 2 moles Pb we need 1 mol O2 and 4 moles H+ to produce 2 moles H2O and 2 moles Pb^2+

For 0.00130 moles Pb^2+ produced we need 0.00130 moles Pb

Step 5: Calculate mass of Pb

Mass Pb = moles Pb * molar mass Pb

Mass Pb = 0.00130 moles * 207.2 g/mol

Mass Pb = 0.269 grams

0.269 grams of lead metal reacted

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marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago

What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O?

yawa3891 [41]

The balanced chemical reaction will be:

2H2O = 2H2 + O2

We are given the amount of water used in the decomposition reaction. This will be our starting point.

17.0 g H2O (1 mol H2O/ 18.02 g H2O) (1 mol O2/2 mol H2O) ( 32.00 g O2/1mol O2) = 15.09 g O2

Percent yield = actual yield / theoretical yield x 100

Percent yield =10.2 g / 15.09 g x 100

Percent yield = 67.58%

5 0