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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 in the reaction between chemical A and B, you need the balanced chemical equation, the molar masses of A, B, and the product, and the stoichiometry of the reaction. Then, convert the masses of reactants into moles, identify the limiting reactant, calculate the theoretical yield of the product using stoichiometry, and finally convert the moles of the product to grams to determine the amount of product produced.

Step by step solution

01

Write the balanced chemical equation for the reaction between A and B

It's important to know the balanced equation representing the chemical reaction between A and B to understand the stoichiometry or mole ratios of reactants and products involved in the reaction. For the purpose of explanation, let's assume the balanced chemical equation is: A + B -> C
02

Determine the molar masses of A, B, and the product C

To convert the given masses of reactants A and B into moles, we need to know their molar masses. We also need to know the molar mass of the product C to express the amount of product formed in grams. Suppose the molar masses of A, B, and C are: \(M_A\), \(M_B\), and \(M_C\)
03

Calculate the moles of reactants A and B

Using the molar masses of A and B, convert the given mass of reactants into moles using the formula: Moles of A = \(\frac{mass~of~A}{molar~mass~of~A} = \frac{10.0~g}{M_A}\) Moles of B = \(\frac{mass~of~B}{molar~mass~of~B} = \frac{10.0~g}{M_B}\)
04

Identify the limiting reactant

The limiting reactant is the reactant that is completely consumed in a chemical reaction and determines the amount of product that can be formed. Compare the mole ratios of A and B in the balanced equation with the calculated moles of A and B. The reactant with the smaller mole ratio is the limiting reactant.
05

Calculate the theoretical yield of product C using stoichiometry

Since we now know the limiting reactant, we can use stoichiometry to determine the theoretical yield of product C. According to the balanced equation, 1 mole of A reacts with 1 mole of B to form 1 mole of C. Theoretical yield of C (moles) = Moles of limiting reactant (either A or B, whichever is smaller)
06

Convert the moles of product C to grams

In order to find the amount of product formed in grams, use the molar mass of product C: Amount of product C (grams) = Moles of product C * Molar mass of C This final value gives the amount of product C that will be produced from the given amounts of reactants A and B.

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Most popular questions from this chapter

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of \(0.157 \mathrm{~g}\) of the compound produced \(0.213 \mathrm{~g}\) \(\mathrm{CO}\), and \(0.0310 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that \(0.103 \mathrm{~g}\) of the compound produces \(0.0230 \mathrm{~g} \mathrm{NH}_{3} .\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

A sample of urea contains \(1.121 \mathrm{~g} \mathrm{~N}, 0.161 \mathrm{~g} \mathrm{H}, 0.480 \mathrm{~g} \mathrm{C}\), and \(0.640 \mathrm{~g} \mathrm{O} .\) What is the empirical formula of urea?

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 \(\underline{X}\)

Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \% \mathrm{H} \text { (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \text { II } & 12.58 & 87.42 \\ \text { III } & 2.34 & 97.66 \end{array} $$ When \(1.00 \mathrm{~L}\) of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$ \begin{array}{lcc} & \mathrm{H}_{2} \text { (L) } & \mathrm{N}_{2} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \text { II } & 2.00 & 1.00 \\ \text { III } & 0.50 & 1.50 \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

Balance the following equations: a. \(\mathrm{Cr}(s)+\mathrm{S}_{\mathrm{s}}(s) \rightarrow \mathrm{Cr}_{2} \mathrm{~S}_{3}(s)\) b. \(\mathrm{NaHCO}_{3}(s) \stackrel{\text { Heat }}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{KClO}_{3}(s) \stackrel{\text { Heat }}{\longrightarrow} \mathrm{KCl}(s)+\mathrm{O}_{2}(g)\) d. \(\operatorname{Eu}(s)+\mathrm{HF}(g) \rightarrow \operatorname{EuF}_{3}(s)+\mathrm{H}_{2}(g)\)

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