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The addition of 3.15 g of $\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}$ to a solution of 1.52 $\mathrm{g}$ of $\mathrm{NH}_{4} \mathrm{SCN}$ in 100 $\mathrm{g}$ of water in a calorimeter caused the temperature to fall by $3.1^{\circ} \mathrm{C}$ . Assuming the specific heat of the solution and products is 4.20 $\mathrm{Jg}^{\circ} \mathrm{C}$ , calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:$\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+2 \mathrm{NH}_{4} \mathrm{O} \mathrm{CN}(a q) \rightarrow \mathrm{Ba}(\mathrm{SCN})_{2}(a q)+2 \mathrm{NH}_{3}(a q)+10 \mathrm{H}_{2} \mathrm{O}(l)$

$q=1318.84 \mathrm{J}$

Chemistry 101

Chapter 5

Thermochemistry

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So we're gonna try to figure out how much heat is absorbed by the reaction when we dissolve or when we react 3.15 grams with barium. I dropped signed octo hydrate 3.15 grams and 1.52 grams. Ah, ammonium Cyrus SIA nicked in 100 grams of water. So first we have to figure out our limiting reactions to take a look at our chemical reaction we have for everyone. Molecule of barium, octo hydrate. We have we reacted with two moles Ammonium thigh Asai Innate. So let's convert thes grams into moles. So too do we just have to divide by our muller masses, which I've given you here. So this is going to give us 3.152 Right, abi 3 15.46 is going to give us about 0.1 knows of barium hydroxide and and then 1.52 divided by 76.1 to 2. It is going to give us about 0.2 moons of ammonium Tyrus, I in it. So looking at this, that means that there is no living reaction that we because our chemical reaction is the 1 to 2 ratio and our reactions under 1 to 2 ratio that all of our barium hydroxide octo hydrate is going to be used up when reacting with points here. Two moles of ammonium five Cy in eight. So now we didn't calculate how much energy the reaction is going to absorb. So in this case, que reaction is the equal negative que water. Which means that all of the heat that the water releases is gonna be absorbed by our reaction. So then we just replace it with MSL titty, so key reaction equals. So we have our math. It's gonna be 100. What grams of water plus 3.15 last 1.52 Because those are the masses of our reactant ce. This is the heat is gonna be 4.2, which is what our problem tells us. And then Delta T is gonna be negative 3.1, because again, that's what our problem tells us that the temperature decreases by 3.1, which is where this negative comes from. So if I plug all of these numbers in, so I'm adding these together and that's gonna give me 104.67 Then you multiply it by these other values that's gonna give me que reaction. So I'm gonna read it over here, Q reaction. If I do this expression out is gonna give me 1362 0.8, Jules. And so we see that the sign is positive because our two negative signs cancel out. Which makes sense that this is an endo thermic reaction so that the number should be above zero. Thanks for watching.

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