Factores de conversión y estequiometría

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Warning notice For the following procedures, please notice that the numbers I got may differ from your numbers because of the rounding. Particularly, I worked with all significant figures and rounded at the very end of the final calculation.
Oxalic acid LD (lethal dose= LD50) in rats is 375 mg/kg. - How many grams of it would be toxic for a 160 lb person? A= 27 grams BUT WHY?

There areseveral ways to work this problem. Let me show you one of them (the one I consider to be the easiest and that uses conversion factors): The very first thing you wanna do is to convert the letal dose from mg of poison for each kg of weight to mg of poison for each pound (lb) of weight. Remember one nice hint is to always write the description of the units and not only the units themselves. Thus:
375 1 12.205 ℎ ℎ = 170.07 1





This is the initial data of the mortal dose

This is a conversion factor that you can find in many places (for example, Table 2.9 in page 36 from the T&T)

This is another way to express the mortal dose, but it’s totally equivalent to the first one!

Now you wanna see how many mg of poison you need to kill a 160 lb person. Using conversion factors you’llhave the next expression:
(160 ℎ ) 170.07 1 = 27 210.88



This is the input (the quantity you want to convert)

This is a conversion factor because is the relationship between two quantities

This is the mg of poison you will need for your input

You can also get to the same result by making a proportion (which is another way to solve the problem… whatever works for you, its valid!):170.07 mg of poison x 1 lb of weight 160 lb of weight

1

From where

=

(

.

)(

)

= 27 210.88

Last thing is to notice that 27 thousand mg is a very huge amount, but you can express it as a smaller amount if you change from mg to g. Next I’ll show you the conversion using conversion factors, but you might as well use a proportion (regla de tres):
(27 210.88 ) 1 1000 = 27.211Of course you round the number to only 27.

In an experiment, the mass of a piece of copper id determined to be 8.56 g. Then the copper is reacted wth sufficient oygen gas to produce solid copper ( II ) oxide. 2Cu + O2 = 2CuO hOW MANY GRAMS OF OXYGEN ARE REQUIRED TO COMPLETELY REACT WITH COPPER ? a= 9.47 G WHY? HOW MANY G OF COPPER LL RESULTFOR THE REACTION OF 8.56 GR. OF CUPPER AND 3.72 OFOXYGEN?

For your first question: Let me just warn you that there is an important mistake in the book: the answer for this question (problem CI.13, question e, page 323) is wrong. Since you were the one to find it, how about if you write to the editorial house and correct the mistake? (I’ll help you out with this). Now that you are more relaxed, let me show you how to work the problem: The veryfirst things to do is to notice that you have a quantity of copper that is expressed in grams, and to remember that the coefficients in a reaction only work for moles, not for grams. So you need to transform 8.56 g of copper to moles of copper by using the molar mass in your periodic table. You can carry out a proportion or do it with conversion factors (it is really up to you) I’ll do it here witha proportion, only to follow the same procedure I’ve been doing in the classroom: 1mol of Cu x 63.55 g of Cu 8.56 g of Cu

2

From where

=

( .

)( .

)

= 0.135

Now that you have the amount of Cu expressed in moles, the next thing to do is to see the reaction and notice that every 2 mol of Cu will react with only 1 mol of O2. But you do not have 2 mol of Cu, but only 0.135mol. As you can guess by now, you need another proportion: 2 mol of Cu 0.135 mol of Cu
( . )(

1 mol of O2 x
)

From where

=

= 0.067

Finally, you have to convert moles of O2 to grams of O2. One more time this is done with a simple proportion, using the data from your periodic table. When doing this, you have to be very careful, because in a molecule of O2 you have two atoms of...
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