/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 176 Determine the weights of \(\math... [FREE SOLUTION] | 91Ó°ÊÓ

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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

Expert verified
Upon burning 104 g of Câ‚‚Hâ‚‚, we obtain 352 g of COâ‚‚ and 72 g of Hâ‚‚O.

Step by step solution

01

Calculate the molecular weight of Câ‚‚Hâ‚‚

First, we need to find the molecular weight of Câ‚‚Hâ‚‚. 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 Câ‚‚Hâ‚‚. Therefore, the molecular weight of Câ‚‚Hâ‚‚ is: \[2 \times 12 (C) + 2 \times 1 (H) = 24 + 2 = 26\]
02

Convert grams of Câ‚‚Hâ‚‚ to moles of Câ‚‚Hâ‚‚

Next, convert 104 g of Câ‚‚Hâ‚‚ to moles using the molecular weight we calculated in step 1 (26 g/mol): \[\text{moles of } C_2H_2 = \frac{104 \text{ g}}{26 \frac{\text{g}}{\text{mol}}} = 4 \text{ mol}\]
03

Use stoichiometry to find moles of COâ‚‚ and Hâ‚‚O produced

According to the balanced chemical equation given, for every 2 moles of Câ‚‚Hâ‚‚ burned, 4 moles of COâ‚‚ and 2 moles of Hâ‚‚O are produced. We have 4 moles of Câ‚‚Hâ‚‚, and we need to find the corresponding moles of COâ‚‚ and Hâ‚‚O produced: \[\text{moles of } CO_2 = \frac{4 \text{ moles of } C_2H_2}{2} \times 4 = 8 \text{ moles}\] \[\text{moles of } H_2O = \frac{4 \text{ moles of } C_2H_2}{2} \times 2 = 4 \text{ moles}\]
04

Convert moles of COâ‚‚ and Hâ‚‚O to grams

We're given the molecular weights of COâ‚‚ (44 g/mol) and Hâ‚‚O (18 g/mol). Now, we can convert moles of COâ‚‚ and Hâ‚‚O to grams: \[\text{grams of } CO_2 = 8 \text{ moles} \times 44 \frac{\text{g}}{\text{mol}} = 352 \text{ g}\] \[\text{grams of } H_2O = 4 \text{ moles} \times 18 \frac{\text{g}}{\text{mol}} = 72 \text{ g}\]
05

Write the final answer

Upon burning 104 g of Câ‚‚Hâ‚‚, we obtain 352 g of COâ‚‚ and 72 g of Hâ‚‚O.

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Key Concepts

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

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.

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Most popular questions from this chapter

Balance the following by filling in missing species and proper coefficient: (a) \(\mathrm{NaOH}+\ldots \mathrm{NaHSO}_{4}+\mathrm{HOH}\) (b) \(\mathrm{PCT} 3+\mathrm{HOH} \rightarrow \ldots+3 \mathrm{HCl}\) (c) \(\mathrm{CH}_{4}+\underline{\mathrm{CCl}_{4}+4 \mathrm{HCl}}\)

Two atoms of scandium are to combine with three atoms of oxygen. If you start with 1 gram of scandium, how much oxygen is required? Scandium has an atomic weight of \(44.96 \mathrm{~g} /\) mole. The at. wt of oxygen is \(15.999 \mathrm{~g} / \mathrm{mole}\).

How many moles of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) can be formed when a mixture of \(0.36\) moles of aluminum and \(0.36\) moles of oxygen is ignited? Which substance and how much of it is in excess of that required? \(4 \mathrm{Al}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}\)

Balance the equations: (a) \(\mathrm{Ag}_{2} \mathrm{O} \rightarrow \mathrm{Ag}+\mathrm{O}_{2}\) (b) \(\mathrm{Zn}+\mathrm{HCl}+\mathrm{ZnCl}_{2}+\mathrm{H}_{2}\); (c) \(\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\).

"Hard" water contains small amounts of the salts calcium bicarbonate \(\left(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\right)\) and calcium sulfate \(\left(\mathrm{CaSO}_{4}\right.\), molecular weight \(=136 \mathrm{~g} / \mathrm{mole}\) ). These react with soap before it has a chance to lather, which is responsible for its cleansing ability. \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) is removed by boiling to form insoluble \(\mathrm{CaCO}_{3} \cdot \mathrm{CaSO}_{4}\), is removed by reaction with washing soda \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right.\), molecular weight \(\left.=106 \mathrm{~g} / \mathrm{mole}\right)\) according to the following equation: \(\mathrm{CaSO}_{4}+\mathrm{Na}_{2} \mathrm{CO}_{3} \rightarrow \mathrm{CaCO}_{3}+\mathrm{Na}_{2} \mathrm{SO}_{4}\) If the rivers surrounding New York City have a CaSO \(_{4}\) concentration of \(1.8 \times 10^{-3} \mathrm{~g}\) liter, how much \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is required to "soften" (remove \(\mathrm{CaSO}_{4}\) ) the water consumed by the city in one day (about \(6.8 \times 10^{9}\) liters)?

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