hsc question chem (1 Viewer)

[ta1g]

can someone show me how to work this out

Ozone reacts with nitric oxide according to the equation
NO(g) + O3(g) → NO2(g) + O2(g)
0.66 g NO(g) was mixed with 0.72 g O3(g).
What is the maximum volume of NO2(g) produced at 0°C and 100 kPa?
(A) 0.34 L
(B) 0.37 L
(C) 0.45 L
(D) 0.50 L

 

[pLuvia]

well since a mod told me this before. Year 11 preliminary forum is only restricted to the year 11 things, so post it on the hsc chemistry forum

 

[mitsui]

can someone show me how to work this out

Ozone reacts with nitric oxide according to the equation
NO(g) + O3(g) → NO2(g) + O2(g)
0.66 g NO(g) was mixed with 0.72 g O3(g).
What is the maximum volume of NO2(g) produced at 0°C and 100 kPa?
(A) 0.34 L
(B) 0.37 L
(C) 0.45 L
(D) 0.50 L

lolz. we mte as well practice. since it is not that confusing
1. work out how many moles of NO there is in 0.66g
2. work out how many volumes there is of NO
3. mole ratio NO:NO2 is 1:1
4. the number of volumes of NO reacted is the same as the volumes of NO2 produced.

is that rite? (sorry, i dont have a caculator atm, so yea.. =P)

 

[Skywalker]

1. Work out the number of moles of NO and O3 involved in the reaction (0.022 and 0.015 respectively in a 1:1 ratio)
2. Use the lower figure from the calculation above - this is your limiting factor
3. Hence multiply the number of moles (0.015) by the volume of gas produced at STP (in this case 22.71 litres) - giving you 0.34 litres

 

[mitsui]

1. Work out the number of moles of NO and O3 involved in the reaction (0.022 and 0.015 respectively in a 1:1 ratio)
2. Use the lower figure from the calculation above - this is your limiting factor
3. Hence multiply the number of moles (0.015) by the volume of gas produced at STP (in this case 22.71 litres) - giving you 0.34 litres

oh crap. i forgot the O3 ><..
so we use the limiting reagent and work out the maximum volumes of the NO2 created?