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miskamm [114]
3 years ago
13

Using the Law of Conservation of Matter, determine the number of grams of iron sulfide (FeS) that will be produced in this react

ion 112g Fe + 64g S → _____ g FeS. A. 48 grams B. 64 grams C. 112 grams D. 176 grams
Chemistry
2 answers:
vazorg [7]3 years ago
4 0
The Law of Conservation of Mass states that the mass of reactants entering a reaction must be equal to the mass of the products exiting it. In this case, we only have 2 reactants, Fe and S, and we only have 1 product, FeS. Therefore we expect the total mass of the Fe and S reactants to equal the mass of FeS. This gives us 112 g + 64 g = 176 g of FeS, which is choice D.

Alexxx [7]3 years ago
4 0
In the <span>Law of Conservation of Matter, the mass in should be equal to the mass out. So, for the reaction, 

</span>112g Fe + 64g S →<span>_____ g FeS

Mass in is equal to 112 g plus 64 g which is 176 g. Therefore, the correct answer among the choices is option D.</span>
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Calculate the empirical formula for each natural flavor based on its elemental mass percent composition.
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The complete question is;
Calculate the empirical formula for each of the following naturalflavors based on their elemental mass percent composition.

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mass                             58.80 g                  9.87 g                   31.33
molar mass                   12 g/mol                 1 g/mol                 16 g/mol
number of moles           58.80/12                9.87/1                    31.33/16
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then divide number of moles by least number of moles - 1.95 in this case
                                      4.9/1.95 = 2.51      9.87/1.95 = 5.06    1.95/1.95= 1
next multiply by 2 to get numbers that can be rounded off to whole numbers
                                       2.51x2 = 5.02        5.06x2 = 10.12      1x2 = 2
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 O - 2
therefore empirical formula is C₅H₁₀O₂

Q2) for this too since elemental composition has been given in percentages lets calculate for 100 g of compound 
                                          C                         H                         O
mass                               63.15 g                5.30 g                 31.55 g
molar mass                     12 g/mol              1 g/mol                16 g/mol
number of moles             63.15/12             5.30/1                  31.55/16
                                        =5.26                  =5.30                   =1.97
divide the number of moles by the least number of moles - 1.97
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multiply each by 3 to get numbers that can be rounded off to whole numbers
                                        2.67x3 = 8.01     2.69x3 = 8.07         1x3 = 3
rounded off to the nearest whole numbers 
C - 8
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Im pretty sure that it is true
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