need help with chem please?

Question

400mL H(2) and 400mL O(2) are mixed together and ignited.

a) write a balanced equation

b) what volume is left unreated

c) what volume of water wapor is formed if all gases are measured at 150Celsius?

explain please, im in dire need of help understanding this. thank you for your help. I will pick the best answer right away

Answer 1

1:
2H(2)+1O(2)=2H2O (I hate no sub scripts)

2: First calculate out the number of mol's of each gas, (If we assure normal operating conditions 25c and 1 ATM pressure and ideal performance) pV=nRT, you will find that all the H2 is reacted but only 200ml of the O2 is reacted, (as seen in the balanced chem equation above). So you are left with 200ml of O2 and 400ml of H20 (as you end up with 2 waters for every O2 and 2 H2 you react). The amount of gas measured in mols stays the same, you can't create or destroy matter only change it and since you combined 1 O2 and 2 H2 you get 2 H20 not 3 H20, the volume changes but the total amount of each gas doesn't resultant volume is 600ml

If we heat the gas up we change T in pV=nRT (ideal gas law equation) It all depends on what else is allowed to change, we will assume constant p and n (number of moles of gas). As such we have V2/V1=T2/T1, so using Kelvin temp scale, 423.15k/298.15k*600ml=851.55 ml but given your significant digits above, (assuming 3 sig figs) you end up with 850 ml.

Answer 2

a)2H2 + O2 -> 2H2O

b) Since half of the O2 is unreacted (all of the H2 reacts w/ half of the O2) then 200 ml is unreacted.

c) Approx 200 ml of water. However, that will be dependent on the temp. What was the temp of the original gas? Assuming they were at room temp (20 C=293 K) use temp in Kelvin (150 C=423 K) and you'll get (423/293)*200 ml= 288 ml of H2O at this higher temp.

Answer 3

Looong time since I did this...lol

Answer 4

Try homework help.

Answer 5

try the science and math section, then chemistry

Answer 6

you'll do better in "homework help"