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Lithopone is a brilliant white pigment used in waterbased interior paints. It is a mixture of \(\mathrm{BaSO}_{4}\) and \(\mathrm{ZnS}\) produced by the reaction $$\mathrm{BaS}(\mathrm{aq})+\mathrm{ZnSO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{ZnS}(\mathrm{s})+\mathrm{BaSO}_{4}(\mathrm{s})$$ How many grams of lithopone are produced in the reaction of \(315 \mathrm{mL}\) of \(0.275 \mathrm{M} \mathrm{ZnSO}_{4}\) and \(285 \mathrm{mL}\) of 0.315 M BaS?

Short Answer

Expert verified
The total mass of Lithopone produced in the reaction is 28.66 grams.

Step by step solution

01

Determine the moles of reactants

First we need to determine the amount of the reactants in moles. For this we multiply the volume of each reactant in liters with their corresponding molarity given as \(M\). Therefore, For \(ZnSO_{4}\): \(0.315L\) * \(0.275M\) = \(0.086625 mol\)For \(BaS\): \(285mL = 0.285L\), \(0.285L\) * \(0.315M\) = \(0.089775 mol\)
02

Identify the Limiting Reactant

Since the reaction uses one mole of each reactant to produce one mole of each product, the limiting reactant is \(ZnSO_{4}\) because it has fewer moles available, \(0.086625 mol\), compared to \(BaS\) which has \(0.089775 mol\). Hence, only \(0.086625 mol\) each of \(ZnS\) and \(BaSO_{4}\) can be formed.
03

Convert Moles of product to Grams

Finally, convert moles of chemical products to grams using their molar masses. For \(ZnS\): \(0.086625 mol\) * \(97.47 g/mol = 8.44g\)For \(BaSO_{4}\): \(0.086625 mol\) * \(233.4 g/mol = 20.22g\)
04

Calculate Total Mass of Lithopone

As Lithopone is the mixture of the two products, we would have to add up the masses of \(ZnS\) and \(BaSO_{4}\) which gives us \(8.44g + 20.22g = 28.66g\)

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Key Concepts

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

Limiting Reactant
In chemical reactions, especially those involving multiple reactants, the limiting reactant plays a crucial role. The limiting reactant is the substance that is entirely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed.
Think of it as a bottleneck in the reaction process.
Even if other reactants are present in abundance, the reaction cannot proceed once the limiting reactant is depleted. To identify the limiting reactant, you must compare the mole ratio of the reactants used in the chemical equation to the moles of each reactant you actually have.
  • Calculate moles for each reactant using the formula: Moles = Molarity \(\times\) Volume in liters.
  • Compare the mole ratio based on the balanced chemical equation given.
  • The reactant that produces the least amount of product is your limiting reactant.
This concept is essential not just for stoichiometry, but also for optimizing reactions in industrial applications and laboratory experiments.
Molar Mass
Molar mass is a fundamental concept in chemistry that helps you convert between the mass of a substance and the amount in moles. To determine molar mass, sum up the atomic masses of all atoms in a molecule as listed in the periodic table.
This property enables you to transition seamlessly between grams and moles, hence it's frequently used in stoichiometry exercises.
For example, to find the molar mass of a compound like \(ZnS\) you need:
  • Atomic mass of Zinc \(Zn\), approximately 65.38 g/mol
  • Atomic mass of Sulfur \(S\), approximately 32.07 g/mol
  • Add them together to get \(97.45 g/mol\)
Understanding molar mass allows you to determine how much of a reactant or product is involved in a reaction by providing a way to convert from mass to moles, which aligns perfectly with the concept of stoichiometry.
Chemical Reactions
Chemical reactions are transformations where the reactants undergo a chemical change to form new products. In the context of stoichiometry, chemical reactions are expressed using chemical equations that represent this transformation. These equations are composed of reactants on the left side and products on the right, separated by an arrow indicating the direction of the reaction.
Each chemical reaction requires balancing to ensure that the number of atoms for each element is conserved according to the Law of Conservation of Mass. This means that the mass of the reactants must equal the mass of the products, making stoichiometry exercises a valuable tool for understanding the extent of reactions.
  • Reactants are substances initially present in a reaction.
  • Products are substances produced as a result of the reaction.
  • Stoichiometry relies on the mole concept to correlate the proportions of reactants and products.
Whether you are producing a brilliant white pigment or optimizing a reaction, understanding these principles is fundamental for interpreting chemical equations and predicting the mass and yield of products.

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Most popular questions from this chapter

Iron ore is impure \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) When \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is heated with an excess of carbon (coke), metallic iron and carbon monoxide gas are produced. From a sample of ore weighing \(938 \mathrm{kg}, 523 \mathrm{kg}\) of pure iron is obtained. What is the mass percent \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) by mass, in the ore sample, assuming that none of the impurities contain Fe?

A seawater sample has a density of \(1.03 \mathrm{g} / \mathrm{mL}\) and \(2.8 \% \mathrm{NaCl}\)l by mass. A saturated solution of \(\mathrm{NaCl}\) in water is \(5.45 \mathrm{M} \mathrm{NaCl} .\) How many liters of water would have to be evaporated from \(1.00 \times 10^{6} \mathrm{L}\) of the seawater before \(\mathrm{NaCl}\) would begin to crystallize? (A saturated solution contains the maximum amount of dissolved solute possible.)

A sulfide of iron, containing \(36.5 \%\) S by mass, is heated in \(\mathrm{O}_{2}(\mathrm{g}),\) and the products are sulfur dioxide and an oxide of iron containing \(27.6 \%\) O, by mass. Write a balanced chemical equation for this reaction.

A method for eliminating oxides of nitrogen (e.g., \(\mathrm{NO}_{2}\) ) from automobile exhaust gases is to pass the exhaust gases over solid cyanuric acid, \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) When the hot exhaust gases come in contact with cyanuric acid, solid \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) decomposes into isocyanic acid vapor, HNCO(g), which then reacts with \(\mathrm{NO}_{2}\) in the exhaust gases to give \(\mathrm{N}_{2}, \mathrm{CO}_{2^{\prime}}\) and \(\mathrm{H}_{2} \mathrm{O}\) How many grams of \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) are needed per gram of \(\mathrm{NO}_{2}\) in this method? [Hint: To balance the equation for reaction between HNCO and \(\mathrm{NO}_{2}\), balance with respect to each kind of atom in this order: \(\mathrm{H}, \mathrm{C}, \mathrm{O}, \text { and } \mathrm{N} .]\)

A 0.3126 g sample of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) requires 26.21 mL of a particular concentration of \(\mathrm{NaOH}(\mathrm{aq})\) to complete the following reaction. What is the molarity of the \(\mathrm{NaOH}(\mathrm{aq}) ?\) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

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