Chemistry, 18.10.2019 03:30 ejfleck3655

6. write the ice chart for the reaction of 32.0 g of sulfur and 71.0 g of chlorine: s8 + 4 cl24s2cl2 after completing the chart give a. the mass of product b. the mass of agent in excess c. the mass produced if there is an 87.% yield

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Answer from: dylanjones6996

ICE Table Figure

a. 67.37 g S_2Cl_2

b. 35.62 g Cl_2

c. 58.61 g S_2Cl_2

Explanation:

For the ICE table we have to keep in mind that we have 4 moles of Cl_2 and 1 mol of S_8 and the reactives are consumed, so for Cl_2 we will have -4X and for S_8 we will have -X. Follow the same logic we will have -4X for S_2Cl_2 .

a. Mass of the product

Molar mass of S_8 = 256.52 g/mol

Molar mass of Cl_2 =70.9 g/mol

Molar mass of S_2Cl_2 =135.03 g/mol

We have to find the limiting reagent in the reaction:

S_8+4Cl_2-4S_2Cl_2

$\frac{32}{256.52}=0.124molS_8$

$\frac{71}{70.9}=1 molCl_2$

Divide by the coefficients in the balanced reaction:

$\frac{0.124}{1}=0.124mol$

$\frac{1}{4}=0.25 mol$

The limiting reagent would be S_8

Now is posible to calculate the amount of S_2Cl_2 produced:

$0.124molS_8\frac{4molS_2Cl_2}{1 molS_8}\frac{135.03gS_2Cl_2}{1 molS_2Cl_2}=67.37gS_2Cl_2$

b. Mass in excess

$0.124molS_8\frac{4molCl_2}{1 molS_8}\frac{70.9gCl_2}{1 molCl_2}=35.38gCl_2$

$Excess\hspace{0.1cm}=\hspace{0.1cm}71gCl_2-35.38gCl_2=35.62 gCl_2$

C. 87%Yield

$67.37gS_2Cl_2\frac{87}{100}=58.61gS_2Cl_2$

6. write the ice chart for the reaction of 32.0 g of sulfur and 71.0 g of chlorine: s8 + 4 cl24s2cl

Answer from: Quest

Answer: abeexplanation:

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

Answer from: Quest

Ithink that the answer is b