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1 answer:
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Answer:
1) Fe = 69.9%
O = 31.1%
2) H = 5.19%
O = 16.5%
N = 28.9%
C = 49.5%
Explanation:
One easy way to do percent compositions is to assume you have 100g of a substance.
1) Lets say we have 100g of Fe2O3.
The total molar mass would be:
The molar mass of the Fe2 alone is:
Thus, the grams of Fe2(out of a 100) could be calculated by multiplying 100g * the molar mass ratio of Fe2 to the whole:
Which is approximately 69.9%.
We can find the amount of O3 by simply subtracting, as the rest of the compound is made of O3. Thus, the % composition of O3 is 31.1%
You can then do this same process to the next question, getting us the following:
H = 5.19%
O = 16.5%
N = 28.9%
C = 49.5%
Answer: See attached image
Explanation:
86 percent is the percent yield for this experiment if he expected to produce 5g of product.
Explanation:
Given that:
mass of test tube = 5 grams
mass of test tube + reactant is 12.5 grams
mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)
mass of reactant = 12.5 -5
= 7.5 grams
when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.
so mass of product formed = 9.3 - 5
= 4. 3 grams of product is formed (actual yield)
However, he expected the product to be 5 grams (theoretical yield)
Percent yield = x 100
putting the values in the formula:
percent yield = x 100
= 86 %
86 percent is the percent yield.