2H2 + O2 → 2H2O
First, we will find the moles :
n H2 = m H2 / Mr H2
n H2 = 10 / 2
n H2 = 5 mole
n O2 = m O2 / Mr O2
n O2 = 5 / 16
n O2 = 0.3125 mole
n H2 / coef. H2 > n O2 / coef. O2
So, O2 is the limiting reactant
The mass of water produced :
n H2O = (coef. H2O / coef. O2) • n O2
n H2O = (2/1) • 0.3125
n H2O = 0.625 mole
m H2O = n H2O • Mr H2O
m H2O = 0.625 • 18
m H2O = 11.25 gr
Excesses reactant :
n H2 = 5 - 0.625 = 4.375 mole
m H2 = n H2 • Mr H2
m H2 = 4.375 • 2
m H2 = 8.75 gr
Answer: The final temperature of the gas is 7.58 °C.
Explanation: We are given initial and final pressure of the system and we need to find the final temperature of the system.
To calculate it, we use the equation given by Gay-Lussac.
His law states that pressure is directly related to the temperature of the gas.

Or,

where,
= initial pressure = 893 mmHg = 1.175atm (Conversion factor: 1atm = 760mmHg)
= initial temperature = 49.3°C = [49.3 + 273.15]K = 322.45K
= Final pressure = 778mmHg = 1.023atm
= Final temperature = ?°C
Putting values in above equation, we get:

Converting Final temperature from kelvin to degree Celsius.
![T_2=280.73K=[280.73-273.15]^oC=7.58^oC](https://tex.z-dn.net/?f=T_2%3D280.73K%3D%5B280.73-273.15%5D%5EoC%3D7.58%5EoC)
Hence, the final temperature of the gas is 7.58 °C.
Nonmetals and negative. Answer is c
Answer:


Explanation:
<u>First mixture</u>:
40 wt% methanol - 60 wt% water 200 kg


<u>Second mixture</u>:
70 wt% methanol - 30 wt% water 150 kg


Final mixture:




If, the compositions are constant, the only variables are the mass of each mixture used in the final one, so there can be only one independent balance.