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pav-90 [236]
3 years ago
7

Describe how the mass of the product can be calculated when one reactant is in excess

Chemistry
1 answer:
kvasek [131]3 years ago
3 0
On the off chance that one of the reactants is in overabundance yet you don't know which one it is, you have to compute the hypothetical item mass for the both reactants, with a similar item, and whichever has the lower yield is the one you use to precisely depict masses/sums for the condition, since you can't have more than the non-abundance reactant can create.
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What are some limitations of this experiment? How could it be improved?
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Answer: Limitation: They may be more expensive and time consuming than lab experiments. Limitation: There is no control over extraneous variables that might bias the results. This makes it difficult for another researcher to replicate the study in exactly the same way.

Explanation:

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Lipase is a protein that helps the body break down fats in foods. Lipase is best classified as which type of protein? an enzyme
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Which of the following would make a solution?
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Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou
Ulleksa [173]

Answer:

4.36~g~XY

Explanation:

In this case, we can start with the reaction:

2X + Y_2~->~2XY

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:

3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X

4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2

Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):

\frac{0.04~mol~X}{1}=~0.04

\frac{0.0875~mol~Y_2}{2}=0.04375

The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol Y_2 = 48 g Y_2 (therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY

I hope it helps!

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