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A quantity of \(0.225 \mathrm{~g}\) of a metal \(\mathrm{M}\) (molar mass \(=\) \(27.0 \mathrm{~g} / \mathrm{mol}\) ) liberated \(0.303 \mathrm{~L}\) of molecular hydrogen (measured at \(17^{\circ} \mathrm{C}\) and \(741 \mathrm{mmHg}\) ) from an excess of hydrochloric acid. Deduce from these data the corresponding equation and write formulas for the oxide and sulfate of \(\mathrm{M}\).

Short Answer

Expert verified
The stoichiometric equation for the reaction is \(M + 3HCl -> 1.5H2 + MCl3\). The formula for the oxide of M is \(M2O3\) and the sulfate of M is \(M2(SO4)3\).

Step by step solution

01

Convert mass of Metal into Moles

The molar mass of metal M is given as \(27.0 \mathrm{~g} / \mathrm{mol}\). Therefore, \(0.225 \mathrm{~g}\) of metal M is equivalent to \(0.225/27 = 0.00833 \mathrm{~mole}\) of M.
02

Convert volume of Hydrogen into Moles

Given, the volume of hydrogen gas liberated is \(0.303 \mathrm{~L}\) at \(17^{\circ} \mathrm{C}\) and \(741 \mathrm{mmHg}\). Using the ideal gas law, \(PV=nRT\), with P as the pressure, V as the volume, n is the moles, R is the gas constant and T is the temperature. We first convert the temperature to Kelvin by adding 273 to the Celsius temperature. So, \(17^{\circ} \mathrm{C} = 290 \mathrm{K}\) and the pressure is converted to atm by dividing by 760mmHg/atm. So, \(741 \mathrm{mmHg} = 0.975 \mathrm{atm}\). Substituting P = 0.975, V = 0.303, R = 0.082 (in proper units) and T = 290 gives \(0.975*0.303=n*0.082*290\). Solving for n gives approximately \(n = 0.0120 \mathrm{~mol}\).
03

Deduce the Reaction

From the moles calculated in steps 1 and 2, for every 1 mole of metal M, there are 0.0120/0.00833, approximately 1.4 or nearly 1.5, moles of hydrogen gas H2. Considering the reaction as \(M + xHCl -> x/2H2 + MClx\), the reaction balances best when M is 1 and x is 3 as HCl has surplus amount. The balanced reaction becomes \(M + 3HCl -> 1.5H2 + MCl3\).
04

Write formulas for the Oxide and Sulfate of M

The formulas for the oxide and sulfate of M can be written by considering the valency of M which seems to be 3 (as from the previous step). Therefore, the formula for the oxide of M would be \(M2O3\) (Metal Oxide), and the sulfate of M would be \(M2(SO4)3\) (Metal Sulfate).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental concept in chemistry that relates the pressure, volume, and temperature of a gas to the number of moles. This law is expressed as \(PV = nRT\), where \(P\) represents the pressure of the gas, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant (approximately \(0.0821\) when dealing with atmosphere, liters, and moles), and \(T\) is the temperature in Kelvin.

In the given problem, we use the Ideal Gas Law to calculate the number of moles of hydrogen gas liberated. Firstly, the temperature of hydrogen is converted from Celsius to Kelvin by adding 273, leading to a total of \(290 \, K\).
The pressure is converted from millimeters of mercury (\(741 \, mmHg\)) to atmospheres, yielding \(0.975 \, atm\) (as \(760 \, mmHg\) equals \(1 \, atm\)).

Substituting these values into the Ideal Gas Law formula alongside the known volume of the gas (\(0.303 \, L\)), the equation becomes:
  • \( 0.975 \times 0.303 = n \times 0.082 \times 290 \)
Solving this, we determine that the number of moles of hydrogen gas, \(n\), is approximately \(0.0120 \, mol\).

This calculation is crucial for understanding the relationship between the measured quantities of gases and their corresponding chemical reactions.
Molar Mass Calculations
Molar mass calculations are essential in chemistry for converting between the mass of a substance and the amount in moles. Molar mass is defined as the mass of one mole of a given substance and is expressed in grams per mole \( (\mathrm{g/mol}) \).

In the exercise, we are given the molar mass of the metal \( \mathrm{M} \), which is \(27.0 \, \mathrm{g}/ \mathrm{mol}\). Using the provided mass of \(0.225 \, \mathrm{g}\), we calculate the number of moles with the formula:
  • \( \text{moles} = \frac{\text{mass in grams}}{\text{molar mass}} \)
  • So, \(\frac{0.225}{27.0} = 0.00833 \, \text{mol} \)
These calculations allow us to understand how much of a metal is participating in a reaction. By knowing the moles of metal and the hydrogen gas produced, we can deduce the stoichiometry of the reaction occurring. Such conversions from mass to moles are vital in balancing chemical equations and for making predictions about the outcomes of reactions.
Chemical Formulas
Chemical formulas represent the types and numbers of atoms in a chemical compound, revealing a compound's composition. They are derived by understanding the bonding and valency of the constituent elements.

In our case, we determined that the reaction involving metal \(\mathrm{M}\) and hydrochloric acid \(\mathrm{HCl}\) likely forms \( \mathrm{MCl}_3 \), indicating that metal \( \mathrm{M} \) has a valency of \(3\).
  • Given this valency, we also deduce the formulas for other compounds formed by this metal, such as its oxide and sulfate.
For instance, the oxide of \(\mathrm{M}\) would be \(\mathrm{M}_2\mathrm{O}_3\). Oxides usually consist of a metal and oxygen, with the formula balanced according to their valencies.
Similarly, the sulfate of \(\mathrm{M}\) would be \(\mathrm{M}_2(\mathrm{SO}_4)_3\), where each sulfate ion contributes a valency of \(2\), further demonstrating the need for balancing based on the valency of \(\mathrm{M}\).

Understanding these formulas helps in predicting the properties of the compounds and their reactivity, essential for efficient chemistry learning.

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