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Chapter 9: Problem 87

The gaseous hydrocarbon acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2},\) is used in welders' torches because of the large amount of heat released when acetylene burns with oxygen. $$ 2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ How many grams of oxygen gas are needed for the complete combustion of \(150 \mathrm{g}\) of acetylene?

Short Answer

Expert verified
460.8 grams of oxygen gas are needed for the complete combustion of 150 grams of acetylene.

Step by step solution

01

Calculate moles of acetylene

First, we need to convert the mass of acetylene (150g) into moles. We can do this by dividing the mass of acetylene by its molar mass. The molar mass of acetylene (C鈧侶鈧�) is: C = 12.01 g/mol H = 1.01 g/mol Molar mass of C鈧侶鈧� = (2 脳 12.01) + (2 脳 1.01) = 26.04 g/mol Now, we can calculate the moles of acetylene: moles of acetylene = mass of acetylene / molar mass of acetylene moles of acetylene = 150 g / 26.04 g/mol = 5.76 moles
02

Find the molar ratio between acetylene and oxygen

Next, we need to find the molar ratio between acetylene and oxygen from the balanced chemical equation: 2 C鈧侶鈧�(g) + 5 O鈧�(g) 鈫� 4 CO鈧�(g) + 2 H鈧侽(g) The molar ratio between acetylene (C鈧侶鈧�) and oxygen (O鈧�) is 2:5.
03

Calculate moles of oxygen required

Now, we can use the molar ratio to find the moles of oxygen required for the complete combustion of 150g of acetylene: moles of oxygen = (moles of acetylene 脳 5) / 2 moles of oxygen = (5.76 moles 脳 5) / 2 = 14.4 moles
04

Convert moles of oxygen to grams

Finally, we need to convert the moles of oxygen into grams. The molar mass of oxygen (O鈧�) is: O = 16.00 g/mol Molar mass of O鈧� = 2 脳 16.00 = 32.00 g/mol Now, we can calculate the mass of oxygen required: mass of oxygen = moles of oxygen 脳 molar mass of oxygen mass of oxygen = 14.4 moles 脳 32.00 g/mol = 460.8 g In conclusion, 460.8 grams of oxygen gas are needed for the complete combustion of 150 grams of acetylene.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equations
Chemical equations are a way to represent a chemical reaction using symbols and formulas. They provide information about the reactants (substances that start the reaction) and the products (new substances formed after the reaction). For example, in the combustion of acetylene, the chemical equation is:
\[ 2 \mathrm{C}_{2} \mathrm{H}_{2}(g) + 5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g)\]This equation tells us that 2 moles of acetylene react with 5 moles of oxygen to produce 4 moles of carbon dioxide and 2 moles of water.
  • It is important to ensure that chemical equations are balanced. A balanced equation has the same number of each type of atom on both sides of the arrow. This reflects the conservation of mass, as no atoms are lost or gained during a chemical reaction.
  • Coefficients (numbers in front of formulas) are used to balance equations and show the proportion of reactants to products.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). To find the molar mass of a compound, you add up the molar masses of all its atoms. This is a crucial step in converting between mass and moles, which is important in stoichiometry, the calculation of reactants and products in chemical reactions.
In the given problem, the molar mass of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), is calculated as follows:
  • Carbon (C) has a molar mass of 12.01 g/mol.
  • Hydrogen (H) has a molar mass of 1.01 g/mol.
  • Molar mass of \(\mathrm{C}_{2} \mathrm{H}_{2} = (2 \times 12.01) + (2 \times 1.01) = 26.04 \text{ g/mol}\).
This calculated molar mass helps in determining the number of moles in a given mass of a substance. For example, converting 150 grams of acetylene to moles involves dividing the mass by the molar mass, giving 5.76 moles.
Combustion Reactions
Combustion reactions are exothermic reactions, meaning they release heat. They typically occur when a substance combines with oxygen to form oxides.
For instance, in the combustion of acetylene (a hydrocarbon), acetylene reacts with oxygen gas, \(\mathrm{O}_{2}\), to form carbon dioxide, \(\mathrm{CO}_{2}\), and water vapor, \(\mathrm{H}_{2}\mathrm{O}\):
\[2 \mathrm{C}_{2} \mathrm{H}_{2}(g) + 5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g)\]
  • In a combustion reaction like this one, hydrocarbons produce large amounts of energy, making them useful in applications like welding.
  • Understanding the stoichiometry of combustion helps in calculating the amount of oxygen required for complete combustion, as in our problem where 150 grams of acetylene requires a calculated 460.8 grams of oxygen.
These reactions also demonstrate the importance of stoichiometry in determining how much of each reactant is necessary to ensure a reaction proceeds to completion.

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Most popular questions from this chapter

What do the coefficients of a balanced chemical equation tell us about the proportions in which substances react on a macroscopic (mole) basis?

For each of the following unbalanced chemical equations, suppose \(25.0 \mathrm{g}\) of each reactant is taken. Show by calculation which reactant is limiting. Calculate the theoretical yield in grams of the product in boldface. a. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathbf{C} \mathbf{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathbf{N} \mathbf{O}(g)\) c. \(\mathrm{NaClO}_{2}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{ClO}_{2}(g)+\mathbf{N a C l}(a q)\) d. \(\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow \mathbf{N} \mathbf{H}_{3}(g)\)

The compound sodium thiosulfate pentahydrate, \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O},\) is important commercially to the photography business as "hypo," because it has the ability to dissolve unreacted silver salts from photographic film during development. Sodium thiosulfate pentahydrate can be produced by boiling elemental sulfur in an aqueous solution of sodium sulfite. $$\mathrm{S}_{8}(s)+\mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s)$$ (unbalanced) What is the theoretical yield of sodium thiosulfate pentahydrate when \(3.25 \mathrm{g}\) of sulfur is boiled with 13.1 g of sodium sulfite? Sodium thiosulfate pentahydrate is very soluble in water. What is the percent yield of the synthesis if a student doing this experiment is able to isolate (collect) only \(5.26 \mathrm{g}\) of the product?

Before going to lab, a student read in his lab manual that the percent yield for a difficult reaction to be studied was likely to be only \(40 . \%\) of the theoretical yield. The student's prelab stoichiometric calculations predict that the theoretical yield should be 12.5 g. What is the student's actual yield likely to be?

Many metals occur naturally as sulfide compounds; examples include \(\mathrm{ZnS}\) and \(\mathrm{CoS}\). Air pollution often accompanies the processing of these ores, because toxic sulfur dioxide is released as the ore is converted from the sulfide to the oxide by roasting (smelting). For example, consider the unbalanced equation for the roasting reaction for zinc: $$\mathrm{ZnS}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{ZnO}(s)+\mathrm{SO}_{2}(g)$$ How many kilograms of sulfur dioxide are produced when \(1.0 \times 10^{2} \mathrm{kg}\) of \(\mathrm{ZnS}\) is roasted in excess oxygen by this process?
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