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Chapter 5: Problem 53

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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Most popular questions from this chapter

Under constant-pressure conditions a sample of hydrogen gas initially at \(88^{\circ} \mathrm{C}\) and \(9.6 \mathrm{~L}\) is cooled until its final volume is \(3.4 \mathrm{~L}\). What is its final temperature?

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