Reacting masses and volumes (of solutions and gases)

When 50 cm³ of hydrocarbon Y is burnt, it reacts with exactly 300 cm³ of oxygen to form 200 cm³ of carbon dioxide. Water is also formed in the reaction. Deduce the equation for this reaction. Explain your reasoning

Combustion of hydrocarbons typically involves a hydrocarbon (`C_xH_y`) reacting with oxygen (`O_2`) to produce carbon dioxide (`CO_2`) and water (`H_2O`).

We have:

50 cm³ of hydrocarbon Y

300 cm³ of oxygen

200 cm³ of carbon dioxide

The volumes of gases involved in the reaction can be directly related to the stoichiometry of the gas reaction, according to Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain an equal number of molecules. 

Thus, the volume ratio of `O_2` to `CO_2` is 300:200, which simplifies to 3:2

For the general equation:

`C_x​H_y​+O_2​→CO_2​+H_2​O`

From the deduction above, we have 3 volumes of `O_2` reacting to produce 2 volumes of `CO_2`. This implies that for every hydrocarbon molecule burned, three `O_2`molecules are required in order to produce two `CO_2` and water.

Assuming hydrocarbon Y could be ethane (`C_2H_6`), as such

`C_2​H_6​+3O_2​→2CO_2​+3H_2​O`

According to this reaction, one hydrocarbon molecule combines with three oxygen molecules to form two carbon dioxide molecules and three water molecules.

In conclusion, hydrocarbon Y burns to produce water and carbon dioxide while preserving mass conservation.

Give all answers to 3 significant figures.

4.80 dm³ of chlorine gas was reacted with sodium hydroxide solution. The reaction taking place was as follows:

`Cl_2(g) + 2NaOH(aq) ->  NaCl(aq) + NaOCl(aq) + H_2O(l)`

a. How many moles of Cl₂ reacted?

b. What mass of NaOCl was formed?

c. If the concentration of the NaOH was 2.00 mol dm^–3, what volume of sodium hydroxide solution was required?

d. Write an ionic equation for this reaction.

a. We can assume the condition of this reaction is in standard temperature and pressure (STP). Thus,

Number of moles of Cl₂ = `4.80 / 22.4` = 0.214 moles

b. From the balanced chemical equation, 1 mole of Cl₂ produces 1 mole of NaOCl

=> Number of moles of Cl₂ = Number of moles of NaOCl = 0.214 moles

Mass of NaOCl = Molar mass of NaOCl x Number of moles of NaOCl

=`(22.99 + 16.00 + 35.45) xx 0.214` = 15.930 g

c. 2 x Number of moles of Cl₂ = Number of moles of NaOH = `0.214 xx 2` = 0.428 moles

Volume of NaOH solution required = `0.428 / 2.00` = 0.214 dm³

d. The balanced net ionic equation includes only species that are involved in the reaction. It can be found by removing ions that occur on both the reactant and product side of the complete ionic equation.

`Cl_2(g) + 2OH^(-)(aq) -> Cl^(-)(aq) + ClO^(-)(aq) + H_2O(l)`

Ammonia gas combines with hydrogen chloride gas, forming solid ammonium chloride. 

a. Write a balanced chemical equation for this reaction.

b. In a reaction mixture of 3.0 g ammonia and 5.0 g hydrogen chloride, which of the two is the limiting reagent? 

c. How many grams of ammonium chloride could form from the reaction mixture in part b? 

d. How much of the reactant is left over in the reaction mixture?

a. The balanced chemical equation for the reaction between ammonia gas (`NH₃`) and hydrogen chloride gas (`HCl`) to form solid ammonium chloride (`NH₄Cl`) is: `NH_3(g) + HCl (g) → NH_4​Cl(s)`

b. The molar mass of `NH_3` = 17.03 g/mol

Moles of `NH_3` = `3.0 / 17.03` = 0.176 moles

The molar mass of HCl = 36.45 g/mol

Moles of HCl = `5 / 36.45` = 0.137 moles

We have less moles of HCl, hence it is the limiting reagent since the balanced equation indicates that the reaction ratio is 1:1.

c. Thus, the mass of solid ammonium chloride will be calculated following moles of HCl

The molar mass of `NH_4Cl` is 53.49 g/mol

Mass of `NH_4Cl` = `0.137 xx 53.49` = 7.33 g

d. NH3 is left over after the reaction

Remaining moles of `NH_3` = `0.176 - 0.137` = 0.039 moles

Mass of remaining `NH_3` = `0.039 xx 17.03` = 0.664 g

A student started with a 3.51 g piece of magnesium ribbon. They placed the magnesium ribbon inside a crucible and heated the magnesium directly to make magnesium oxide.

a. Calculate the number of moles of magnesium used in the reaction.

b. Calculate the theoretical mass of magnesium oxide that can be produced in the reaction. 

c. The student actually produced 4.11 g of magnesium oxide. Calculate the percentage yield of their reaction.

a. The number of moles of magnesium used in the reaction = `3.51 / 24.00` = 0.14625 moles

b. The theoretical mass of magnesium oxide that can be produced in the reaction = `0.14625 xx 40` = 5.85 g

c. Percentage yield = `"actual mass" / "theoretical mass" xx 100` = `4.11 / 5.85 xx 100` = 70.3 %

i. The fermentation of glucose, C₆H₁₂O₆, is one method scientists can use to produce ethanol. Write a balanced symbol equation for this reaction.

ii. During the fermentation process in the reaction in part (i), 312 g of ethanol was produced from 1.5 kg of glucose. 

Calculate the percentage yield of this reaction to 1 decimal place.

i. `C_6H_12O_6 -> 2C_2H_5OH + 2CO_2`

ii. The molar mass of Glucose = 180.0 g/mol 

Moles of Glucose = `(1.5 xx 1000) / 180.0` = 8.3 moles

Moles of ethanol = `2 xx 8.3` = 16.6 moles

The theoretical mass of ethanol = `16.6 xx 46` = 763.6 g

Percentage yield =  `"actual mass" / "theoretical mass" xx 100` = `312 / 763.6 xx 100` = 40.9 %

Potassium dichromate, K₂Cr₂O₇ is a useful oxidising agent. A laboratory technician wanted to make up enough of this solution for a class practical. 

Calculate the mass of potassium dichromate needed to prepare 3.00 dm³ of a 1.50 x 10^-3 mol dm^-3 solution. 

Give your answer to the appropriate precision.

Moles of K₂Cr₂O₇ = `3.00 xx 1.5 xx 10^(-3) = 4.5 xx10^(-3` moles

The molar mass of K₂Cr₂O₇ = 294 g/mol

Mass of K₂Cr₂O₇ = `294 xx 4.5 xx 10^(-3)` = 1.32 g

A student completes a titration experiment to determine the percentage by mass of iron in steel wire.

Their titration results are shown in the table.

Calculate the mean titre.

A. 24.36 cm³

B. 24.30 cm³

C. 24.28 cm³

D. 24.37 cm³

The answer is C.

Identify the concordant (within 0.1 cm³) results => 24.25 and 24.30

The average of the concordant results = `(24.25 + 24.30)/2` = 24.275 = 24.28 cm³

A is incorrect because this is the average of all four titre which has no concordant results been identified

B is incorrect because this is the average of the three concordant results in the table but the rough titre has been used when it should not be included

D is incorrect because this is the average of titres 1,2,3 which has no concordant results

A student reacted copper metal with excess sulfuric acid and collected 35.9 g of the hydrated copper sulfate crystals, CuSO₄.5H₂O crystals. The yield of crystals was 80.0 %.

What was the original mass of copper used?

A. 7.3 g

B. 9.1 g

C. 11.4 g

D. 17.9 g

The answer is C.

The molar mass of CuSO₄.5H₂O = `63.5 + 32.1 + 4 xx 16.0 + 5 xx (2 xx 1.0 + 16.0)` = 249.6 g

The expected mass from a 100 % yield = `35.9 / 80.0 xx 100` = 44.875 g

Moles of CuSO₄ = `44.875 / 249.6` = 0.180 moles

Using the 1:1 relationship, moles of CuSO₄ = moles of Cu = 0.180 moles

Thus, the original mass of copper = `0.180 xx 63.5` = 11.4 g

What is the minimum volume of 0.5 mol dm^-3 sulfuric acid needed to fully react with 2.43 g of magnesium powder?

`Mg (s) + H_2SO_4 (aq) → MgSO_4 (aq) + H_2 (g)`

A. 20 cm³

B. 50 cm³

C. 100 cm³

D. 200 cm³

The answer is D.

Moles of Mg = `2.43 / 24.3` = 0.1 moles

There are 0.5 moles of sulfuric acid in 1000 cm³. Hence, to obtain the 0.1 moles of sulfuric acid required to react with the 0.1 moles of magnesium = `1000 / 5` = 200 cm³

A volume of nitrogen gas, 3.00 dm³ measured under room temperature conditions, was reacted with a large volume of hydrogen gas, to produce ammonia. Only 15.0 % of the nitrogen gas reacted to produce ammonia.

What mass of ammonia was made? 

A. 6.83 g

B. 1.42 g

C. 0.638 g

D. 0.513 g

The answer is A.

`N_2 + 3H_2 -> 2NH_3`

Moles of Nitrogen gas = `3.00 / 24` = 0.125 moles

Only 15.0 % of the nitrogen used = `0.15 xx 0.125` = 0.01875 moles

The molar mass of NH₃ = `14.0 + 3 xx 1.0` = 17 g/mol

1 moles of nitrogen will produce 2 moles of NH₃

Moles of NH₃ = `2 xx 0.01875` = 0.0375 moles

The mass of ammonia = `17 xx 0.0375` = 0.638 g