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Alexandra [31]
3 years ago
5

What mass of water can be obtained from 4.0 g of H2 and 16 g of O2?

Chemistry
1 answer:
rusak2 [61]3 years ago
4 0

Answer:

18 grams of water, will be produced from 4 g of H₂ and 16g of O₂

Explanation:

Reaction: 2H₂ + O₂  ⇒  2H₂O

Moles = Mass / Molar mass

Molar mass H₂ = 2g/m

Molar mass O₂ = 32 g/m

4g /2g/m = 2moles

16g / 32 g/m = 0.5 moles

Ratio is 2:1

For 2 moles of hydrogen, I need 1 mol of oxygen; I have 2 moles of H₂, but I only have 0.5 of O₂, so the O₂ is the limiting reagent.

1 mol of O₂ produces 2 mol of water

0.5 moles of O₂ will produce, the double → 1 mol of water.

Molar mass of water is 18 g/m

1 mol . 18 g/m = 18 g

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How many liters of CO2 are in 4.76 moles? (at STP)
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<u>Answer:</u> The volume of carbon dioxide gas at STP for given amount is 106.624 L

<u>Explanation:</u>

We are given:

Moles of carbon dioxide = 4.76 moles

<u>At STP:</u>

1 mole of a gas occupies a volume of 22.4 Liters

So, for 4.76 moles of carbon dioxide gas will occupy a volume of = \frac{22.4L}{1mol\times 4.76mol=106.624L

Hence, the volume of carbon dioxide gas at STP for given amount is 106.624 L

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3 years ago
Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha
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Answer:

0.269 g

Explanation:

Po_{210} ^{84}  ⟶  Pb_{206} ^{82}+  + He_{4} ^{2}

Data:

Half-life of  Po_{210} ^{84}  (T(1/2)) = 138.4 days

Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol  

Time = 338.8 days  

Isotopic masses  

Po_{210} ^{84} = 209.98 g/mol  

Pb_{206} ^{82} = 205.97 g/mol  

Concepts  

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.  

The radioactive decay law is  

N=Noe^(-λt)

Where:  

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days  

λ= radioactive decay constant  

The radioactive constant is related to the half-life by the next equation  

λ= \frac{ln 2}{t(1/2)}

so  

λ= \frac{ln2}{138.4 days}  =0,005008 days^(-1)

No (Atoms of  Po_{210} ^{84}  in t=0)

To get No we need to calculate the number of atoms of  Po_{210} ^{84}   in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of  Po_{210} ^{84}  in the initial sample.  

This is:

\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol} = 0,001599 mol of  PoCl4

This is the number of mol of  Po_{210} ^{84} in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23  

0,001599 mol of  Po_{210} ^{84} * 6.023*10^23 atoms/ mol of  Po_{210} ^{84} = 9.632 *10^20 atoms of  Po_{210} ^{84}

Atoms after 338.8 days

We use the radioactive decay law to get this value  

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of  Po_{210} ^{84} in the sample after 338.8 days has passed  

The number of atoms  Po_{210} ^{84} transformed is equal to the number of atoms of Pb_{206} ^{82}  produced.  

The number of atoms of Po_{210} ^{84} transformed is No - N  

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of Pb_{206} ^{82} produced  

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of Pb_{206} ^{82} ) / ( 6.023*10^23 atoms of Pb_{206} ^{82} / mol of Pb_{206} ^{82} =  0.001306 moles of Pb_{206} ^{82}

This number of moles have a mass of:

(0,001306 moles of Pb_{206} ^{82} )* (205.97 g of Pb_{206} ^{82} /mol of Pb_{206} ^{82} ) = 0.269 g  

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