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Slav-nsk [51]
3 years ago
13

How many moles of Fe2O3 are in 209 g of the compound?

Chemistry
1 answer:
Bezzdna [24]3 years ago
7 0
M = 209g
M = 56 * 2 + 16 * 3 = 160 g/mol

n = m/M = 209g / 160 g/mol = 1,30625mol
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What will happen to the wavelength of a wave if the frequency is left unchanged?
algol [13]
Someone. or something will get hurt
6 0
3 years ago
Can someone help me with this molar mass problem?[It’s the last one]
quester [9]

Answer:

54.18 \times 10^{23} \ moles in 3 mole of  Al_2(SO_4)_3

Explanation:

It is clear that in the given 1\ mole of Al_2(SO_4)_3 have 3\ ions of SO_4^2^-

Therefore 3 moles of Al_2(SO_4)_3 will have 3\times3=9 \ ions of   SO_4^2^-

Since 1 ion of anything is equivalent to 6.02\times10^{23} \ moles

Therefore 3 moles of Al_2(SO_4)_3 will have 3\times3=9 \ ions of   SO_4^2^-

Which is equivalent to 9 \times6.02\times10^{23}=54.18\times10^{23} \ moles

Thus 3 moles of  Al_2(SO_4)_3 gives 54.18\times10^{23} \ moles of  SO_4^2^-.

5 0
3 years ago
The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)
cestrela7 [59]

Answer:

a. The limiting reactant is NaHCO_{3}

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

a)

You know the following reaction:

3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}⇒3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

  • Na:  23
  • H: 1
  • C: 12
  • O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

  • NaHCO_{3} :23+1+12+16*3=84 g/mol
  • H_{3} C_{6} HO_{7} :1*3+12*6+1*5+16*7= 192 g/mol
  • CO_{2} :12+16*2= 44 g/mol
  • H_{2} O :1*2+16= 18 g/mol
  • Na_{3} C_{6} H_{5} O_{7} : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of NaHCO_{3} react with 1 mole of H_{3} C_{6} HO_{7}  Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

  • NaHCO_{3} : 252 g
  • H_{3} C_{6} HO_{7} : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:  

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So <u><em>the limiting reagent is sodium bicarbonate</em></u>.

b)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of NaHCO_{3} react with 3 mole of CO_{2}. Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

  • NaHCO_{3} : 252 g
  • H_{3} C_{6} HO_{7} : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}

<u><em>grams of carbon dioxide=  0.73 g</em></u>

<u><em>Then, 0.73 g of carbon dioxide are formed.</em></u>

c)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

grams of citric acid=\frac{1.4 g * 192 g}{252 g}

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

<em><u>So the grams of excess reactant that do not participate in the reaction are 0333 g.</u></em>

3 0
3 years ago
What substance do you place in the burette during a titration
strojnjashka [21]

Answer:

Acid solution

Explanation:

In acid-base titrations carried out in school and college labs, many of the older generation, well-informed teachers told their students that the acid solution should be taken in the burette.

5 0
2 years ago
What should the IUPAC name for a binary covalent compound lack ?
zavuch27 [327]
The answer to the question is "B. Roman Numerals"
5 0
3 years ago
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