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Chapter 8: Problem 99

Equal masses of iron and sulphur are heated together to form FeS. What fraction of the original mass of excess reactant is left unreacted? \((\mathrm{Fe}=56\), \(\mathrm{S}=32\) ) (a) \(0.22\) (b) \(0.43\) (c) \(0.86\) (d) \(0.57\)

Short Answer

Expert verified
Answer: The fraction of the original mass of excess reactant left unreacted is 0.43.

Step by step solution

01

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between iron and sulphur is: \(\mathrm{Fe + S \rightarrow FeS}\)
02

Calculate the moles of each reactant

Assume we have M grams of each reactant. To determine the moles, we'll divide the mass by the molar mass of each respective element: Moles of Fe = \(\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}} = \frac{M}{56}\) Moles of S = \(\frac{\text{Mass of S}}{\text{Molar mass of S}} = \frac{M}{32}\)
03

Determine the limiting reactant

Since the balanced chemical equation tells us that one mole of iron reacts with one mole of sulphur to produce one mole of iron sulfide, we can compare the moles of each reactant to determine which one is the limiting reactant. Moles of Fe: \(\frac{M}{56}\) Moles of S : \(\frac{M}{32}\) Since \(\frac{M}{32}\) > \(\frac{M}{56}\), Fe is the limiting reactant.
04

Calculate the fraction of excess reactant left unreacted

Now that we've determined that Fe is the limiting reactant, we can calculate the fraction of excess sulfur left unreacted. First, calculate the mass of sulphur that reacts with iron: Mass of reacted S = Mass of Fe \(*\) \(\frac{\text{Molar mass of S}}{\text{Molar mass of Fe}}\) Mass of reacted S = \(M \cdot \frac{32}{56}\) Then, calculate the mass of unreacted excess sulfur: Mass of unreacted excess S = Original mass of S - Mass of reacted S Mass of unreacted excess S = M - \(M \cdot \frac{32}{56}\) To calculate the fraction of unreacted excess sulfur, divide the mass of unreacted excess sulfur by the original mass of sulfur: Fraction = \(\frac{\text{Mass of unreacted excess S}}{\text{Original mass of S}}\) Fraction = \(\frac{M - M \cdot \frac{32}{56}}{M}\) Fraction = \(\frac{56-32}{56}\) Fraction = \(\frac{24}{56}\) Fraction = \(0.43\) So the fraction of the original mass of excess reactant left unreacted is 0.43, which corresponds to option (b).

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Key Concepts

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

Stoichiometry
Stoichiometry is the section of chemistry that deals with the proportional relationship between reactants and products in chemical reactions. It is essentially a form of bookkeeping for chemistry, ensuring that atoms are conserved during chemical transformations. To solve stoichiometric problems, one must have a balanced chemical equation, which provides the mole ratios of reactants and products, serving as a conversion factor.

To tackle stoichiometry problems, follow these basic steps:
  • Balance the chemical equation for the reaction.
  • Convert all given information into moles (using molar masses).
  • Use the mole ratios from the balanced equation to calculate the amount of desired product or reactant.
  • Convert moles back into the required units of measurement.
In limiting reactant problems, the stoichiometry allows us to find out which reactant will be completely consumed first and thus limits the amount of product formed.
Chemical Reaction Equations
Chemical reaction equations are symbolic representations of chemical reactions. In these equations, the reactants are listed on the left side while the products are shown on the right, with the arrow indicating the direction of the transformation. A well-balanced chemical equation adheres to the law of conservation of mass, meaning the number of atoms for each element must be the same on both sides of the equation.

For instance, the equation \(\mathrm{Fe + S \rightarrow FeS}\) represents the reaction between iron (Fe) and sulfur (S) to create iron sulfide (FeS). Each element on both sides needs to be counted to ensure mass balance. This provides a foundation for stoichiometric calculations, as it reveals the mole ratio of the reactants and products. It is essential that before solving stoichiometry problems, one always starts with a correctly balanced chemical equation.
Molar Mass Calculations
Molar mass calculations help determine the mass of one mole of a substance, which is pivotal in converting between mass and moles during chemical reactions. Molar mass is measured in grams per mole (g/mol) and can be found on the periodic table as the atomic weight for each element.

To calculate molar masses in a compound, sum up the atomic weights of all the atoms present in the molecule. For example, in the reaction of iron and sulfur forming iron sulfide (FeS), the molar mass of iron (Fe) is 56 g/mol and sulfur (S) is 32 g/mol. These values are crucial when determining the moles of each reactant given a certain mass.
  • Moles of Fe = \(\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}} = \frac{M}{56}\)
  • Moles of S = \(\frac{\text{Mass of S}}{\text{Molar mass of S}} = \frac{M}{32}\)
By carrying out molar mass calculations, one can easily convert the mass of reactants to the mole quantities needed for stoichiometric comparisons in chemical reactions.

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Most popular questions from this chapter

A quantity of \(2.0 \mathrm{~g}\) of a triatomic gaseous element was found to occupy a volume of \(448 \mathrm{ml}\) at \(76 \mathrm{~cm}\) of \(\mathrm{Hg}\) and \(273 \mathrm{~K}\). The mass of its each atom is (a) \(100 \mathrm{amu}\) (b) \(5.53 \times 10^{-23} \mathrm{~g}\) (c) \(33.3 \mathrm{~g}\) (d) \(5.53\) amu

Mole fraction of solute in an aqueous solution of \(\mathrm{NaOH}\) is \(0.1 .\) If the specific gravity of the solution is \(1.4\), the molarity of the solution is (a) \(6.93\) (b) \(0.1\) (c) \(71.4\) (d) \(0.14\)

The empirical formula of an organic gaseous compound containing carbon and hydrogen is \(\mathrm{CH}_{2}\). The volume occupied by certain mass of this gas is exactly half of the volume occupied by the same mass of nitrogen gas under identical conditions. The molecular formula of the organic gas is (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{CH}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{12}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8}\)

The volume of one mole of water at \(277 \mathrm{~K}\) is \(18 \mathrm{ml}\). One \(\mathrm{ml}\) of water contains 20 drops. The number of molecules in one drop of water will be \(\left(N_{\mathrm{A}}=6 \times 10^{23}\right)\) (a) \(1.07 \times 10^{21}\) (b) \(1.67 \times 10^{21}\) (c) \(2.67 \times 10^{21}\) (d) \(1.67 \times 10^{20}\)

In Dumas method, \(0.2 \mathrm{~g}\) of an organic nitrogenous compound gave \(28 \mathrm{ml}\) of \(\mathrm{N}_{2}\) (volume reduced to \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) ). What is the percentage of nitrogen, by mass, in the compound? (a) \(17.5\) (b) \(8.75\) (c) \(35.0\) (d) \(14.0\)
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