Consider the following balanced chemical equation: $$ A+5 B \longrightarrow 3 C+4 D $$ a. Equal masses of \(\mathrm{A}\) and \(\mathrm{B}\) are reacted. Complete each of the following with either "A is the limiting reactant because \(" ;{ }^{\prime *} \mathrm{~B}\) is the limiting reactant because \("\) or "we cannot determine the limiting reactant because i. If the molar mass of \(\mathrm{A}\) is greater than the molar mass of B, then ii. If the molar mass of \(\mathrm{B}\) is greater than the molar mass of A, then b. The products of the reaction are carbon dioxide (C) and water (D). Compound \(\mathrm{A}\) has the same molar mass as carbon dioxide. Compound \(\mathrm{B}\) is a diatomic molecule. Identify \(\mathrm{com}-\) pound \(\mathrm{B}\) and support your answer. c. Compound \(\mathrm{A}\) is a hydrocarbon that is \(81.71 \%\) carbon by mass. Detemine its empirical and molecular formulas.

Step by step solution

01

(Question a, part i) Determine the limiting reactant when the molar mass of A is greater than B.

Since we are given equal masses of A and B, we need to consider their molar ratios to determine the limiting reactant. First, we'll compare the molar mass. Given that the molar mass of A is greater than the molar mass of B, then the number of moles of A in equal masses will be less than the moles of B. Therefore, A is the limiting reactant because we cannot form more products when A is completely consumed.

02

(Question a, part ii) Determine the limiting reactant when the molar mass of B is greater than A.

In the case where the molar mass of B is greater than the molar mass of A, the number of moles of B in equal masses will be less than the moles of A. Therefore, B is the limiting reactant because we cannot form more products when B is completely consumed.

03

(Question b) Identify compound B

In the chemical equation, compound C is carbon dioxide (CO₂), and compound D is water (H₂O). Compound A has the same molar mass as CO₂, which means that the molar mass of A is \(44.01 g/mol\). Compound B is a diatomic molecule. To identify compound B, we'll look at the balanced chemical equation: \(A + 5B \longrightarrow 3C + 4D\). Since A is a hydrocarbon, it will only contain carbon (C) and hydrogen (H). The balanced chemical equation suggests that compound B must be reactant which could provide ample oxygen (O) to form CO₂ and H₂O. So B is diatomic oxygen, O₂.

04

(Question c, part i) Determine the empirical formula of compound A

We are given that compound A is a hydrocarbon and is \(81.71\%\) carbon by mass. Therefore, the remaining mass percentage in compound A is hydrogen, which is \(100\%-81.71\%=18.29\%\). To determine the empirical formula, we'll assume we have 100 grams of the compound, so: - Carbon: \(81.71 g\) (mass) × \(\frac{1 mol}{12.01 g}\) (molar mass) = \(6.808 mol\) C - Hydrogen: \(18.29 g\) (mass) × \(\frac{1 mol}{1.008 g}\) (molar mass) = \(18.148 mol\) H Divide the moles of each element by the smaller of the two mole values: - Carbon: \(\frac{6.808}{6.808}\) = 1 - Hydrogen: \(\frac{18.148}{6.808}\) ≈ 2.67 Since it is close to 3, we can approximate it to get the empirical formula: \(\mathrm{CH_3}\).

05

(Question c, part ii) Determine the molecular formula of compound A

Since compound A has the same molar mass as CO₂, \(44.01 g/mol\), we need to determine the multiple of the empirical formula mass which will match this molar mass. First, let's calculate the empirical formula mass: Empirical formula mass = \(12.01 (C) + 3(1.008 (H)) = 12.01 + 3.024 = 15.034 g/mol\) Now, we'll determine the multiple of the empirical formula mass that matches the molar mass: Molar mass multiple = \(\frac{44.01}{15.034}\) ≈ 2.92, which we can approximate to 3. Therefore, the molecular formula of compound A is 3 times the empirical formula: Molecular formula = \(\mathrm{C_3H_{9}}\)

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