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3 years ago

CaF2 + (NH4)2O --> 2 NH4F + CaO How many grams of NH4F can be formed from 34.6 grams of CaF2? (AKS 4f)

Chemistry

1 answer:

bixtya [17]3 years ago

5 0

Answer:

32.83 grams of NH₄F can be formed from 34.6 grams of CaF₂

Explanation:

The balanced reaction is:

CaF₂ + (NH₄)₂O ⇒ 2 NH₄F + CaO

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities participate in the reaction:

CaF₂: 1 mole
(NH₄)₂O: 1 mole
NH₄F: 2 mole
CaO: 1 mole

Being the molar mass of the compounds:

CaF₂: 78 g/mole
(NH₄)₂O: 52 g/mole
NH₄F: 37 g/mole
CaO: 56 g/mole

Then by stoichiometry, the following amounts of mass participate in the reaction:

CaF₂: 1 mole* 78 g/mole= 78 g
(NH₄)₂O: 1 mole* 52 g/mole= 52 g
NH₄F: 2 mole* 37 g/mole= 74 g
CaO: 1 mole* 56 g/mole= 56 g

You can apply the following rule of three: if 78 grams of CaF₂ form 74 grams of NH₄F by stoichiometry, 34.6 grams of CaF₂ will form how much mass of NH₄F?

$mass of NH_{4} F=\frac{34.6 grams of CaF_{2} *74 grams ofNH_{4} F}{78grams of CaF_{2} }$

mass of NH₄F= 32.83 grams

32.83 grams of NH₄F can be formed from 34.6 grams of CaF₂

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Many computer chips are manufactured from silicon, which occurs in nature as SiO2. When SiO2 is heated to melting, it reacts wit

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Answer:

a) The limiting reagent in this reaction is SiO₂ .

b) The theoretical yield of Si from this reaction = 72,494.85 g = 72.5 kg

c) The percent yield of Si from this reaction = 91%

Explanation:

a) The limiting reagent is the reagent whose amount at the start of the reaction is in shortage according to the stoichiometric balance. It is is the reactant that determines how much of other reactants will react and how much products will be formed. It is theoretically, completely used up in the reaction.

The non-limiting reagent is usually in excess according to the stoichiometric balance.

154.9 kg of SiO2 is allowed to react with 78.0 kg of carbon to produce 66.0 kg of silicon.

The balanced equation for the reaction is

SiO₂ + 2C -------> Si + 2CO

1 mole of SiO₂ reacts with 2 moles of C according to the stoichiometric balance

To obtain which reactant is in excess and which one is the limiting reagent, we have to find the number of moles of reactant present at the start of the reaction.

Number of moles = (mass)/(molar mass)

For SiO₂, mass = 154.9 kg = 154,900 g, Molar mass = 60.02 g/mol

Number of moles = (154900/60.02)

Number of moles = 2580.81 moles

For Carbon, mass = 78.0 kg = 78,000 g, Molar mass = 12.011 g/mol

Number of moles = (78000/12)

Number of moles = 6494.05 moles

Recall, 1 mole of SiO₂ reacts with 2 moles of C

If Carbon was the limiting reagent,

6494.05 moles of Carbon would require (6494.05/2) moles of SiO₂ to react; 3247.025 moles of SiO₂. Which is more than the available number of moles of SiO₂ at the start of the reaction. Hence, Carbon isn't the limiting reagent.

SiO₂ as the limiting reagent,

1 mole of SiO₂ reacts with 2 moles of Carbon,

2580.81 moles of SiO₂ would react with (2×2580.81) moles of Carbon; 5161.62 moles of Carbon. Which is in the limit of available number of moles of Carbon at the start of the reaction. Hence, SiO₂ is the limiting reagent which determines which amount of other reactants react and the amount of products formed.

b) Theoretical yield of Si in the reaction.

SiO₂ + 2C -------> Si + 2CO

SiO₂ being the limiting reagent.

1 mole of SiO₂ gives 1 mole of Si,

2580.81 moles of SiO₂ will give 2580.81 moles of Si.

Mass produced = (number of moles produced) × (Molar mass)

Number of moles of Si produced = 2580.81 moles

Molar mass of Si = 28.09 g/mol

Theoretical mass of Si produced = (2580.81) × (28.09) = 72494.85 g = 72.5 kg

c) Percemt yield of Si

Percent yield = 100% × (Actual yield)/(Theoretical yield)

Actual yield of Si = 66 kg

Theoretical yield of Si = 72.49485 kg

Percent yield = 100% × (66/72.49485)

Percent yield = 91.04% = 91%

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To solve this problem, first we assume the volume is purely additive. The density of the mixture can then be calculated by the summation of mass fraction of each component divided by its individual density:

1 / ρ mixture = (x NH3 / ρ NH3) + (x H2O / ρ H2O) ---> 1

Calculating for mass fraction of NH3:

x NH3 = 15 g / (15 g + 250 g)

x NH3 = 0.0566

Therefore the mass fraction of water is:

x H2O = 1 – x NH3 = 1 – 0.0566

x H2O = 0.9434

Assuming that the density of water is 1 g / mL and substituting the known values back to equation 1:

1 / 0.974 g / mL = [0.0566 / (ρ NH3)] + [0.9434 / (1 g / mL)]

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Given the density of NH3, now we can calculate for the volume of NH3:

V NH3 = 15 g / 0.680 g / mL

V NH3 = 22.07 mL

The number of moles NH3 is: (molar mass NH3 is 17.03 g / mol)

n NH3 = 15 g / 17.03 g / mol

n NH3 = 0.881 mol

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Answer:

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Explanation:

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Molar mass of $BaSO_4$ = 233.43 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

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$Moles= \frac{12.5221\ g}{233.43\ g/mol}$

Moles of $BaSO_4$ = 0.0536 moles

According to the given reaction,

$Ba^{2+}_{(aq)}+SO_4^{2-}_{(aq)}\rightarrow BaSO_4_{(s)}$

1 mole of $BaSO_4$ is formed from 1 mole of $SO_4^{2-}$

Thus,

0.0536 moles of $BaSO_4$ is formed from 0.0536 moles of $SO_4^{2-}$

Moles of $SO_4^{2-}$ = 0.0536 moles

Moles of sulfur in 1 mole $SO_4^{2-}$ = 1 mole

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Answer:

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The given unbalanced chemical equation is:

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From the equation, it can be seen that there are three atoms of hydrogen on the products side, however, only two on the reactant's side. So, in order to balance the equation, one needs to multiply the sodium hydroxide by 2 to get a total of 4 atoms of hydrogen on the product's side.

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