Question

Determine the weights of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced on burning \(104 \mathrm{~g}\). of \(\mathrm{C}_{2} \mathrm{H}_{2}\). Molecular weights are \(\mathrm{CO}_{2}=44\) and \(\mathrm{H}_{2} \mathrm{O}=18\). The equation for the reaction is \(2 \mathrm{C}_{2} \mathrm{H}_{2}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)

Short answer

Upon burning 104 g of C₂H₂, we obtain 352 g of CO₂ and 72 g of H₂O.

Step-by-step solution

Calculate the molecular weight of C₂H₂

First, we need to find the molecular weight of C₂H₂. The molecular weight of carbon (C) is 12, and the molecular weight of hydrogen (H) is 1. There are 2 carbon atoms and 2 hydrogen atoms in C₂H₂. Therefore, the molecular weight of C₂H₂ is: \[2 \times 12 (C) + 2 \times 1 (H) = 24 + 2 = 26\]

Convert grams of C₂H₂ to moles of C₂H₂

Next, convert 104 g of C₂H₂ to moles using the molecular weight we calculated in step 1 (26 g/mol): \[\text{moles of } C_2H_2 = \frac{104 \text{ g}}{26 \frac{\text{g}}{\text{mol}}} = 4 \text{ mol}\]

Use stoichiometry to find moles of CO₂ and H₂O produced

According to the balanced chemical equation given, for every 2 moles of C₂H₂ burned, 4 moles of CO₂ and 2 moles of H₂O are produced. We have 4 moles of C₂H₂, and we need to find the corresponding moles of CO₂ and H₂O produced: \[\text{moles of } CO_2 = \frac{4 \text{ moles of } C_2H_2}{2} \times 4 = 8 \text{ moles}\] \[\text{moles of } H_2O = \frac{4 \text{ moles of } C_2H_2}{2} \times 2 = 4 \text{ moles}\]

Convert moles of CO₂ and H₂O to grams

We're given the molecular weights of CO₂ (44 g/mol) and H₂O (18 g/mol). Now, we can convert moles of CO₂ and H₂O to grams: \[\text{grams of } CO_2 = 8 \text{ moles} \times 44 \frac{\text{g}}{\text{mol}} = 352 \text{ g}\] \[\text{grams of } H_2O = 4 \text{ moles} \times 18 \frac{\text{g}}{\text{mol}} = 72 \text{ g}\]

Write the final answer

Upon burning 104 g of C₂H₂, we obtain 352 g of CO₂ and 72 g of H₂O.

Key concepts

Molecular Weight

Understanding molecular weight is fundamental in stoichiometry as it allows us to convert between grams and moles of a substance. Molecular weight, also known as molar mass, is the mass of one mole of a chemical compound. It is expressed in grams per mole (g/mol). To determine the molecular weight of a compound, you add up the atomic weights of all the atoms in its formula. For example, in the compound acetylene (\(\mathrm{C}_2\mathrm{H}_2\)), we have two carbon atoms and two hydrogen atoms.

• The atomic weight of carbon (C) is 12 g/mol.
• The atomic weight of hydrogen (H) is 1 g/mol.

Thus, the molecular weight of \(\mathrm{C}_2\mathrm{H}_2\) is calculated as follows: \[2 \times 12 + 2 \times 1 = 26\]This means that one mole of \(\mathrm{C}_2\mathrm{H}_2\) weighs 26 grams. Knowing this allows you to convert a given mass in grams into the number of moles, which is crucial for understanding how much of each reactant is required or how much product will be formed in a chemical reaction.

Balanced Chemical Equation

A balanced chemical equation is essential in stoichiometry because it provides the relationship between the quantities of reactants and products. The coefficients in a balanced equation indicate the molar ratios of each substance involved in the reaction.
Consider the combustion of acetylene, \(\mathrm{C}_2\mathrm{H}_2\), which reacts with oxygen:
\[2 \mathrm{C}_2\mathrm{H}_2 + 5 \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2 + 2 \mathrm{H}_2\mathrm{O}\]
This equation tells us that:

• 2 moles of \(\mathrm{C}_2\mathrm{H}_2\) react with 5 moles of \(\mathrm{O}_2\).
• The reaction produces 4 moles of \(\mathrm{CO}_2\) and 2 moles of \(\mathrm{H}_2\mathrm{O}\).

These relationships are critical for calculating how much product can be formed from given amounts of reactants or how much of a reactant is needed to completely react with another. The balancing of equations ensures the law of conservation of mass is obeyed, meaning the mass of the reactants equals the mass of the products.

Conversion of Grams to Moles

Converting between grams and moles is a vital part of many stoichiometric calculations. This conversion is typically done using the molecular weight of a compound. Knowing the conversion is like having a roadmap for understanding chemical reactions at a quantitative level.
For example, to convert 104 grams of \(\mathrm{C}_2\mathrm{H}_2\) to moles, given that its molecular weight is 26 g/mol, you would set up the conversion as follows:
\[\text{moles of } \mathrm{C}_2\mathrm{H}_2 = \frac{104 \text{ grams}}{26 \text{ g/mol}} = 4 \text{ moles}\]This calculation is important because it translates a measurable amount, the mass, into a unit of the quantity of substance, moles.
From here, you can use the balanced chemical equation to see that these 4 moles of \(\mathrm{C}_2\mathrm{H}_2\) produce 8 moles of \(\mathrm{CO}_2\) and 4 moles of \(\mathrm{H}_2\mathrm{O}\). Finally, by converting these moles back to grams, you get a tangible result about how much of each product is formed in the reaction.