A group 3A metal has a density of 2.70 gcm3 and a cubic unit cel...

Question

A group 3A metal has a density of 2.70 g>cm3 and a cubic unit cell with an edge length of 404 pm. Reaction of a 1.07 cm3 chunk of the metal with an excess of hydrochloric acid gives a colorless gas that occupies 4.00 L at 23.0 °C and a pressure of 740 mm Hg.
(a) Identify the metal.

Accepted Answer

Hi everyone. This problem reads the density of a group to a metal is 1. g per cubic centimeter and crystallizes in a cubic unit cell. That has an edge length of 653 PICO meters when a 1.65 cc slab of this metal reacts with excess hydro bronek acid. The gas produced occupies a volume of 1.50 liters with the pressure of 794 millimeters of mercury at 27 degrees Celsius. What is the identity of the metal? So, the question that we want to answer is the identity of the metal. Okay. And we know that is it's a group to a medal. All right. So, we're going to say that our group to a medal is going to be represented by M Okay. And we also have hydro bro mc acid. So, let's go ahead and disassociate this into its ions. So we have the hydro knee um, ion plus bromide and ion. Okay. And we know that the metal is reacting with excess hydro bronek acid. So let's go ahead and write an equation for this. So, we have a metal is going to react and we're going to have two moles of hydro of hydro ni um this yields are metal cat ion plus hydrogen gas. Alright, so the first thing that we want to do is we want to solve for the mass of the metal. Alright, so we want to know the mass of the metal and we can calculate the mass of the metal using the density given. So the density Of the metal we're told we have when a 1.65 cm slab of this metal. So that's the mass. So we have 1. cubic centimeters or that's not the mask. Excuse me. That is the volume. Okay. And so what we're going to do is we're going to use the density of the metal that's given to solve for the mass. Okay, So we're going to say we're going to set the density up as a conversion. Okay, so we're told that the density is 1.55 g per cubic centimeter. Alright, so we have 1.55 g per cubic centimeter is the density of the metal. So let's go ahead and cross out cubic centimeters and we're left with grams. So the mass of the metal is 2. g. Okay, so now let's go ahead and convert our pressure two. We're going to convert our pressure to Atmospheres. So let's go ahead and write that. Step two is conversions. Alright, so we have our Pressure is equal to 794 mm of mercury. And we want to go from mm of mercury to atmospheres. Okay, so our conversion is in one atmosphere there is 760 of mercury and mm of Mercury cancel. And are left with atmospheres. So once we do this calculation we get 1. 47 atmospheres. Okay, now we need to convert our temperature. So our temperature is given in degrees Celsius. So we have 27 degrees Celsius and we want to convert this to kelvin. So to do that, we're going to add 273.15. So our temperature in kelvin ends up being 300. kelvin. Alright. And now we can go ahead and use the ideal gas law. So three are ideal gas law is PV equals N. R. T. So we want to solve for the moles of hydrogen gas. Okay, so we need to rearrange this equation so that we're solving for moles. So we'll go ahead and divide both sides by R. T. And so we're going to get N. Is equal to P. V over R. T. Okay, so let's go ahead and plug in what we know we know that P is pressure and we said that the pressure that we converted is 1.447 atmospheres B is volume and the volume was given in the problem. 1.50 leaders are is gas constant a value we should know 0.8 to 06 Leaders atmosphere over mole kelvin And temperature we converted is 300.15 Kelvin. Okay, so we can go ahead and do this calculation. And when we do the calculation we get the moles of hydrogen gas is equal to zero moles hydrogen gas. All right. So now that we know the moles of hydrogen gas, let's look at the mole to mole ratio between our metal and our hydrogen gas. So let's go look up at our reaction that we wrote. So we have our metal and hydrogen gas. So we have a 1 to 1 mole ratio. Alright, so we can write down here, we have a 1 to 1 mole ratio and this is between the metal and hydrogen gas. Okay, so what that means is our moles for the metal is going to equal 0.06362 moles. Alright, so now we can calculate the molar mass because molar masses grams per mole. We know what the masses and the problem because they were told the mass is, Let me see the mass is two point 5575 g. Okay, that was the mass of the metal that we calculated in the first step. Alright, so we're going to figure out the molar mass. Remember molar mass is grams per mole. Okay, so moller mass is equal to grams Permal. We calculated the grams. We calculated the grams as the first part of the problem and we said that it is 2.5575 g. And we just calculated the moles 0.6362 moles. So our molar mass Is equal to 40.2 g per mole. When we look at the periodic table closest in the periodic table to this molar mass is calcium. So our answer is going to be calcium. Okay, because we're able to compare that to the molar mass, so the identity of the metal is calcium and that is it for this problem, I hope this was helpful.

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