Question

I will post the questions in the attachments

Answer 1

@Michele_laino

Answer 2

hint:
here the ratio moles of nitrogen over moles of ammonia is \(1:2\), the same goes for volume at STP conditions

Answer 3

@Rushwr can you see if you can help

Answer 4

What do u think the answer is ?

Answer 5

I think that the answer is C because it seems to make sence

Answer 6

Yeah that's right :)

Answer 7

The molar ratio = the volume ratio

Answer 8

@michele_laino do you agree with me

Answer 9

yes! you are correct! It is option C, namely \(V=22.4\) liters

Answer 10

I think that this one is A

Answer 11

first step:
how many moles of glucose we have?

Answer 12

950 grams?

Answer 13

that is the starting mass of glucose, so the moles of glucose are:
\[n = \frac{{950}}{{180}} = ...?\]
\(180\) being the molecular weight of glucose

Answer 14

ok I am following you so far

Answer 15

please what is \(950/180=...?\)

Answer 16

5.27

Answer 17

that is what you get when you divide those two numbers

Answer 18

correct! we have \(5.278\) moles of glucose.
Now, looking at your chemical reaction, we have \(6\) moles of consumed oxygen for every one mole of glucose, so your reaction happens, if the moles of oxygen are:
\[{n_1} = 5.278 \cdot 6 = ...?\]

Answer 19

that would equal 31.668

Answer 20

correct!

Answer 21

no what do we do

Answer 22

finally, being \(16\) grams the weigh of one mole of oxygen, then the requested quantity of oxygen which has to be consumed is:
\[m = 31.668 \cdot 16 = ...?\]

Answer 23

weight*

Answer 24

that would equal 506.668

Answer 25

please wait, we have to consider the molecular weight of the molecular oxygen, so the correct quantity of oxygen is:
\[m = 31.668 \cdot 32 = ...?\]
since the molecular weight of molecular oxygen is \(16 \cdot 2=32\)

Answer 26

do you want the answer to that?

Answer 27

what is \(31.668 \cdot 32=...?\)

Answer 28

1013.376

Answer 29

so, what is the right option?

Answer 30

C

Answer 31

correct!

Answer 32

can you help me with three more

Answer 33

ok! Please close this question and open a new one