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2 years ago

How many oxygen atoms are contained in 2.74 g of Al2(SO4)3?

Chemistry

2 answers:

skad [1K]2 years ago

4 0

Answer:

either 12 or 24

Explanation:

abruzzese [7]2 years ago

3 0

Answer:

5.79 × 10^23 Oxygen atoms

Explanation:

Number of Oxygen atom in the compound = 4×3 = 12

Molar mass of Al2(SO4)3 = 342 g/mol.

No of mole = mass/molar mass = 2.74/342 = 8.01×10^-03 mole

2.74g of Al2(SO4)3 × 1 mole of Al2 (SO4)3 / 342g of Al2 (SO4)3 * 12 mole of Oxygen/ 1mole of Al(SO4)3 * 6.02×10^23 Oxygen atom/ 1 mole of Oxygen

= 5.79×10^23 Oxygen atoms

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Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

galina1969 [7]

Answer:

$\%\, yield\, \, SO_3=82.29$

Explanation:

First write the balance eqation of chemical reaction:

$2S(s) +3O_2(g) \rightarrow 2SO_3(g)$

Remember writing any chemical reaction from its elemetal form then write the elements in their natural form i.e. how that element exists in the nature here sulphur exists in solid monoatomic form in the nature and oxygen in gaseous diatomic form.

mass of oxygen given=5gram

mole of oxygen $=5/32mol=0.16mol$

mass of sulpher given=6gram

mole of sulpher $=6/32mol=0.19mol$

from the above balanced equaion;

2 mole of sulphur reacts with 3 mole Oxygen completely

1 mole of sulphur reacts with 1.5 mole Oxygen completely

0.19 mole of sulphur reacts with $1.5\times 0.19$ i.e. 0.285 mole Oxygen completely.

but we have 0.16 mole so oxygen will be the limiting reagent and sulpher will be the excess reagent

so product will depend on the limiting reagent

from the balance equation

3 mole of sulpher gives 2 mole $SO_3$

1 mole of sulpher will give 2/3 mole $SO_3$

0.16 of sulpher will give 0.12 mole $SO_3$

mass of $SO_3$ =9.6gram this is theoreical production of $SO_3$

and

actual production of $SO_3$ =7.9gram

$\%\, yield \,\, SO_3=\frac{Actual \,yield}{theoretical\, yield}\times 100$

$\%\, yield\, \, SO_3=\frac{7.9}{9.6} \times 100=82.29$

$\%\, yield\, \, SO_3=82.29$

3 0

2 years ago

Question 2 of 5<br> What do you know about a substance it you know its temperature?

notka56 [123]

One may know how close the molecules within the substances are packed together. Hot substances have molecules that are farther apart, cold substances have molecules that are more compact/closer together.

5 0

3 years ago

Read 2 more answers

How many milliliters of 0.125 M FeCl3 are needed to react with an excess of Na2S to produce 3.75 g of Fe2S3 if the percent yield

Katyanochek1 [597]

Answer:

0.912 mL

Explanation:

3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)

FeCl3 is the limiting reactant.

Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles

Hence actual yield of Iron III sulphide = 0.043 moles

Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield

Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide

From the reaction equation,

2moles of iron III chloride produced 1 mole of iron III sulphide

x moles of iron III chloride, will produce 0.057 of iron III sulphide

x= 2× 0.057= 0.114 moles of iron III chloride

But

Volume= number of moles/ concentration

Volume= 0.114/0.125

Volume= 0.912 mL

4 0

3 years ago

How much CuSO4*5H2O and Na2CO3 is needed to obtain 5g of CuCO3?

Svetllana [295]

I think the reaction that is occurring is CuSO₄(aq)+Na₂CO₃(aq)⇒CuCO₃(s)+Na₂SO₄(aq). Water in the hydrate will just become part of water in solution so it does not really matter for this question.
SInce we now know the balanced chemical equation, you can use stoichiometry to find the amount of reactants needed to produce 5g CuCO₃.
The first step to any stoichiometry question is to convert the mass given into moles. To do this you have to divide the mass by the molar mass of the compound the mass is referring to. The molar mass of CuCO₃ is 123.5g/mol so you have to divide 5g by 123.5g/mol to get 0.04047 moles of CuCO₃.
The next step in this qustoin is to find the number of moles of each reactant using the moles of product we found in the first step. To do this, we need go back and look at the equation to find the molar ratios which is shown through the coefficients. Since all of the coeficients in the chemical equation is 1, we know the molar ratios are all `1 to 1 and therefore the number of moels of each reactant is equal to the number of moles of the product found in step one. This means that you started off with 0.04047 moles of both CuSO₄ and Na₂CO₃.
The final step is to multiply the number moles of each reactant by its molar mass. the molar mass of CuSO₄ is 159.6g/mol and the molar mass of Na₂CO₃ is 106g/mol. When you multiply 0.04047mol by 159.6g/mol you get 6.4g and when you multiply 0.04047mol by 106g/mol you get 4.29g.
Therefore you started with 6.4g of CuSO₄ and 4.29g of Na₂CO₃.
I hope this helps.

5 0

3 years ago

How do you convert scientific notation into SI base units?

Jobisdone [24]

For example 4.698*10^4
This is scientific notation
=46980

6 0

3 years ago