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Calculate the number of grams of HI that are at equilibrium with 1.25 $\mathrm{mol}$ of $\mathrm{H}_{2}$ and 63.5 $\mathrm{g}$ of iodine at 448$^{\circ} \mathrm{C} .$$\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI} \quad K_{c}=50.2 \mathrm{at} 448^{\circ} \mathrm{C}$

507 $\mathrm{g}$

Chemistry 102

Chapter 13

Fundamental Equilibrium Concepts

Chemical Equilibrium

Aqueous Equilibria

Carleton College

University of Kentucky

University of Toronto

Lectures

00:41

In chemistry, an ion is an atom or molecule that has a non-zero net electric charge. The name was coined by John Dalton for ions in 1808, and later expanded to include molecules in 1834.

10:03

In thermodynamics, a state of thermodynamic equilibrium is a state in which a system is in thermal equilibrium with its surroundings. A system in thermodynamic equilibrium is in thermal equilibrium, mechanical equilibrium, electrical equilibrium, and chemical equilibrium. A system is in equilibrium when it is in thermal equilibrium with its surroundings.

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Calculate the number of gr…

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Calculate the number of mo…

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Consider the following equ…

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At a certain temperature, …

So here we've got a reaction vessel that equilibrium that contains hydrogen gas. I didn't gas and hydrogen gas. And what we're asked to do is take two concentrations of you known concentrations Mahajan Gas and I didn't gas and use those to solve for the concentration of the 100. Tonight, I'd gas. So the first thing we actually have to do your is were provided the amount of iodine in grams. We need to first convert that to moles. So we're gonna take the gram amount. We're gonna use a conversion factor where one more bite and gas is 253 18 grams. So we divide and we find that there's about 0.25 moles of iodine in this container. And even though we don't know the exact volume of the container, we can use the molar values directly because the ratio between hydrogen gas night and gas is 1 to 1. So now we're just going to take our equilibrium expression, which, in this case, Casey is equal to hydrogen iodide about it. Her jing gas. I'm dying. We're gonna play these values into this equation. So if we rewrite everything here, we have 50.2 is equal to the concentration of H I, where divided by 1.25 at this 0.25 If we rearrange this to solve for 100 I'd I'd we find that this is equivalent to the square root 50.2. That was 1.25 just 0.25 And from this, we find that the concentration, partisan iodide, equilibrium or the Moler amount here is gonna be 3.96 moles. And if we multiply that by the molar mass of hydrogen iodide, we find that that's equivalent to 507 grams.

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