Question

what mass of water (in grams) forms from the reaction of 2.30 l of hydrogen gas and 1.86 l of oxygen gas? both gases are at 700 torr and 11.1 oc.

Answer 1

The mass of water (in grams) formed from the reaction of 2.30 L of hydrogen gas and 1.86 L of oxygen gas is 1.63 grams

How to calculate the mass of water formed?

First, we shall determine which element is the limiting reactant. Details below:

`2H_2(g)\ +\ O_2(g)\ \rightarrow\ 2H_2O(g)`

From the balanced equation above,

2 L of H₂ reacted with 1 L of O₂

Therefore,

2.30 L of H₂ will react with =
`(2.3\ *\ 1)/(2)` = 1.15 L of O₂

Now, we can see that only 1.15 L of O₂ reacted. Thus, H₂ is the limiting reactant.

Next, we shall calculate the volume of water formed. Details below:

`2H_2(g)\ +\ O_2(g)\ \rightarrow\ 2H_2O(g)`

From the above equation,

2 L of H₂ reacted to produced 2 L of H₂O

Therefore,

2.30 L of H₂ will produced 2.30 L of H₂O

Finally, we shall obtain the mass of the water formed. Details below:

• Volume of water formed (V) = 2.30 L
• Temperature (T₁) = 11.1 °C = 11.1 + 273 = 284.1 K
• Pressure (P) = 700 torr = 700 / 760 = 0.921 atm
• Gas constant (R) = 0.0821 atm.L/Kmol
• Molar mass of water (M) = 18 g/mol
• Mass of water formed (m) =?

`m = (MPV)/(RT) \\\\m = (18\ *\ 0.921\ *\ 2.3)/(0.0821\ *\ 284.1)\\\\m = 1.63\ grams`