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Chapter 3: Problem 6

You know that chemical \(A\) reacts with chemical \(B\). You react \(10.0 \mathrm{~g} A\) with \(10.0 \mathrm{~g} B\). What information do you need to determine the amount of product that will be produced? Explain.

Short Answer

Expert verified
To determine the amount of product produced, you need the balanced chemical equation for the reaction, molar masses of chemicals A and B, and the moles of each reactant. Then, identify the limiting reactant, calculate the moles of the product using the stoichiometric coefficients, and finally, calculate the mass of the product formed using its molar mass.

Step by step solution

01

Identify the balanced chemical equation

Obtain the balanced chemical equation for the reaction between chemical A and B. It should be in the form: \[a A + b B \rightarrow c C\] Here, a, b, and c are the coefficients corresponding to the stoichiometric ratio of the reactants (A and B) and the product (C). These coefficients are crucial for our calculations.
02

Determine the molar masses of chemical A and B

Look up the molar mass of each reactant, chemical A and B, in a periodic table or other reference material. The molar masses will be in units of grams per mole (g/mol).
03

Calculate the moles of each reactant

Divide the mass of each reactant by their respective molar masses to find the moles of each reactant: Moles of A = \( \frac{10.0 \mathrm{~g} A}{Molar\:mass\:of\:A\: (\mathrm{g/mol})}\) Moles of B = \( \frac{10.0 \mathrm{~g} B}{Molar\:mass\:of\:B\: (\mathrm{g/mol})}\)
04

Determine the limiting reactant

Divide the moles of each reactant by their respective coefficients in the balanced chemical equation: Mole ratio of A = \( \frac{Moles\:of\:A}{a}\) Mole ratio of B = \( \frac{Moles\:of\:B}{b}\) The reactant with the lowest mole ratio is the limiting reactant. This determines the maximum amount of product that can be produced.
05

Calculate the moles of product

Use the limiting reactant's mole ratio and the stoichiometric coefficient of the product to calculate the moles of product formed: Moles of product (C) = Mole ratio of limiting reactant × c
06

Calculate the mass of product

Finally, multiply the moles of product by the molar mass of the product to find the mass of product formed: Mass of product (C) = Moles of product (C) × Molar mass of C (g/mol) Now you have determined the amount of product formed in this reaction.

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Most popular questions from this chapter

One of the components that make up common table sugar is fructose, a compound that contains only carbon, hydrogen, and oxygen. Complete combustion of \(1.50 \mathrm{~g}\) of fructose produced \(2.20 \mathrm{~g}\) of carbon dioxide and \(0.900 \mathrm{~g}\) of water. What is the empirical formula of fructose?

An element \(\mathrm{X}\) forms both a dichloride \(\left(\mathrm{XCl}_{2}\right)\) and a tetrachloride \(\left(\mathrm{XCl}_{4}\right)\). Treatment of \(10.00 \mathrm{~g} \mathrm{XCl}_{2}\) with excess chlorine forms \(12.55 \mathrm{~g} \mathrm{XCl}_{4}\). Calculate the atomic mass of \(\mathrm{X}\), and identify \(\mathrm{X}\).

Ascorbic acid, or vitamin \(\mathrm{C}\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\), is an essential vitamin. It cannot be stored by the body and must be present in the diet. What is the molar mass of ascorbic acid? Vitamin C tablets are taken as a dietary supplement. If a typical tablet contains \(500.0 \mathrm{mg}\) vitamin \(\mathrm{C}\), what amount (moles) and what number of molecules of vitamin C does it contain?

Phosphorus can be prepared from calcium phosphate by the following reaction: \(2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+10 \mathrm{C}(s) \longrightarrow\) \(6 \mathrm{CaSiO}_{3}(s)+\mathrm{P}_{4}(s)+10 \mathrm{CO}(g)\) Phosphorite is a mineral that contains \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) plus other non-phosphorus- containing compounds. What is the maximum amount of \(\mathrm{P}_{4}\) that can be produced from \(1.0 \mathrm{~kg}\) of phosphorite if the phorphorite sample is \(75 \% \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) by mass? Assume an excess of the other reactants.

What amount (moles) is represented by each of these samples? a. \(150.0 \mathrm{~g} \mathrm{Fe}_{2} \mathrm{O}_{3}\) b. \(10.0 \mathrm{mg} \mathrm{NO}_{2}\) c. \(1.5 \times 10^{16}\) molecules of \(\mathrm{BF}_{3}\)
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