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Chapter 16: Problem 44

An element ' \(A\) ' reacts with oxygen to produce \(\mathrm{A}_{2} \mathrm{O}_{3}\). \(1 \mathrm{~g}\) of 'A' produces \(1.1596 \mathrm{~g}\) of \(\mathrm{A}_{2} \mathrm{O}_{3}\). Calculate atomic mass of element ' \(A\) ' (Atomic mass of oxygen \(=15.999 \mathrm{~g} \mathrm{~mol}^{-1}\) ): (a) \(20.70\) \(\square\) (b) \(66.85\) [ (c) \(100.2\) \(\square\) (d) \(150.4\)

Short Answer

Expert verified
The atomic mass of element 'A' is approximately 150.4 g/mol.

Step by step solution

01

Identify the Reaction

The element 'A' reacts with oxygen to form the compound \( A_2O_3 \). This indicates that two atoms of A combine with three atoms of oxygen to form one formula unit of \( A_2O_3 \).
02

Determine Mass of Oxygen

Given that 1 g of A produces 1.1596 g of \( A_2O_3 \), we can find the mass of oxygen by subtracting the mass of A from the total mass of the compound: \( 1.1596 \text{ g} - 1 \text{ g} = 0.1596 \text{ g of oxygen} \).
03

Calculate Moles of Oxygen

Given the mass of oxygen is 0.1596 g and its atomic mass is 15.999 g/mol, calculate the moles of oxygen by using: \( \text{moles of oxygen} = \frac{0.1596 \text{ g}}{15.999 \text{ g/mol}} \approx 0.009975 \text{ mol} \).
04

Calculate Moles of \( A_2O_3 \)

Since there are 3 moles of oxygen atoms per mole of \( A_2O_3 \), calculate moles of \( A_2O_3 \) as follows: \( \text{moles of } A_2O_3 = \frac{0.009975 \text{ mol}}{3} = 0.003325 \text{ mol} \).
05

Determine Moles of Element A

The reaction shows that 2 moles of A form 1 mole of \( A_2O_3 \), so: \( \text{moles of A} = 2 \times 0.003325 \approx 0.00665 \text{ mol} \).
06

Calculate Atomic Mass of A

Using the formula \( \text{mass} = \text{molar mass} \times \text{moles} \), solve for molar mass: \( 1 \text{ g} = M \times 0.00665 \text{ mol} \), giving \( M = \frac{1 \text{ g}}{0.00665 \text{ mol}} \approx 150.38 \text{ g/mol} \).
07

Select the Correct Answer

Round the calculated atomic mass of A, 150.38 g/mol, to the nearest choice, which matches \( d) 150.4 \).

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Key Concepts

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

Chemical Reactions
A chemical reaction occurs when substances interact to form new compounds with different properties. In this case, the element 'A' and oxygen react to form the compound \( A_2O_3 \). This is a classic example of a chemical reaction where reactants combine to produce a product. In oxidation reactions, oxygen typically combines with another element to form an oxide. By understanding these basic principles, you gain insight into how materials change and form. Recognizing the empirical formula derived from such interactions allows us to apply stoichiometric principles effectively.
Stoichiometry
Stoichiometry is the field of chemistry that involves calculating the reactants and products in chemical reactions. It is based on the principle of conservation of mass, which states that matter is neither lost nor gained in a chemical reaction.
  • To solve stoichiometry problems, it's crucial to start with a balanced chemical equation.
  • In the exercise, the equation \( 2A + 3O_2 \rightarrow A_2O_3 \) indicates the molar ratios of reactants to products.
  • The stoichiometry allows the determination of how much of each substance will react or be produced.
Understanding stoichiometry helps in calculating the precise amount of required reagents and predicting the outcomes of chemical reactions, as seen in the calculation of the moles of \( A_2O_3 \) and element 'A'.
Oxide Formation
Oxide formation involves a chemical reaction between an element and oxygen, resulting in oxides as products. The compound \( A_2O_3 \) is an example of an oxide formed through this type of reaction. These reactions are typically exothermic, meaning they release energy in the form of heat.
  • Understanding the composition of oxides is critical in identifying the stoichiometry of reactions.
  • This knowledge aids in determining the amount of each substance present in the oxide.
  • In this scenario, we identified that three oxygen atoms combined with two of A to yield \( A_2O_3 \).
Grasping the process of oxide formation is essential for predicting how various elements behave when exposed to oxygen.
Molar Mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It plays a crucial role in the conversion between the mass of a substance and the amount of substance as measured in moles. For example, the atomic mass of oxygen is given as 15.999 g/mol, which is essential in calculating the moles of oxygen present in the compound \( A_2O_3 \). To find the atomic mass of element 'A', the mass of the oxide and the known mass of oxygen were used to determine moles:
  • First, calculate the moles of oxygen using its molar mass.
  • Next, find the moles of \( A_2O_3 \) and subsequently the moles of element 'A'.
  • Finally, use these moles to calculate the atomic mass of 'A'.
By understanding molar mass and its application, solving for unknown quantities in chemical reactions becomes a straightforward task.

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Most popular questions from this chapter

\(2.0 \mathrm{~g}\) mixture of sodium carbonate and sodium bicarbonate on heating to constant weight gave \(224 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) at N.T.P. The \(\%\) weight of sodium bicarbonate in the mixture is: (a) 50 \(\square\) (b) 54 (c) 80 \(\square\) (d) 84

The volume of oxygen at N.T.P. obtained by electrolysis of \(18 \mathrm{~g}\) of water will be : (a) \(22.4\) litre \(\square\) (b) \(5.6\) litre (c) \(11.2\) litre \(\square\) (d) \(44.8\) litre

The mass of \(\mathrm{CaO}\) required to remove the hardness of \(10^{6}\) litre of water containing \(1.62 \mathrm{~g}\) of calcium bicarbonate per litre is : (a) \(140 \mathrm{~kg}\) \(\square\) (b) \(280 \mathrm{~g}\) (c) \(420 \mathrm{~kg}\) \(\square\) (d) \(560 \mathrm{~kg}\)

\(30 \mathrm{~g}\) of magnesium and \(30 \mathrm{~g}\) of oxygen are reacted, then the residual mixture contains : [K.C.E.T. 2002] (a) \(60 \mathrm{~g}\) magnesium oxide only (b) \(40 \mathrm{~g}\) magnesium oxide and \(20 \mathrm{~g}\) of oxygen 1 (c) \(45 \mathrm{~g}\) magnesium oxide and \(15 \mathrm{~g}\) of oxygen . (d) \(50 \mathrm{~g}\) of magnesium oxide and \(10 \mathrm{~g}\) of oxygen \(\square\)

\(7.46 \mathrm{~g}\) of \(\mathrm{KCl}\) are heated in excess of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{MnO}_{2}\). The gas evolved is passed through KI solution. The amount of iodine set free will be : (a) \(12.7 \mathrm{~g}\) \(\square\) (b) \(3.2 \mathrm{~g}\) \(\square\) (c) \(6.4 \mathrm{~g}\) \(\square\) (d) \(18.9 \mathrm{~g}\) \(\square\) [Hint : \(\left[\mathrm{KCl}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{HCl}\right] \times 2\) \(\mathrm{MnO}_{2}+4 \mathrm{HCl} \longrightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) \(2 \mathrm{KI}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{KCl}+\mathrm{I}_{2}\) \(\left.2 \mathrm{KCl}+\mathrm{MnO}_{2}+2 \mathrm{KI} \longrightarrow 2 \mathrm{~K}_{2} \mathrm{SO}_{4}+2 \mathrm{KCl}+\mathrm{MnCl}_{2}+\mathrm{I}_{2}\right]\) \(2 \times 74.5 \mathrm{~g} \quad 2 \times 127 \mathrm{~g}\)
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