When hydrogen selenide gas reacts with oxygen, selenium dioxide and steam are produced.
How many liters of selenium dioxide would be produced from 5.50 L of oxygen? Assume 90% yield and that all gases are measured at the same temperature and pressure.
Balance the equation:
2H2Se + 3O2 > 2SeO2 + 2H2O (g)
For every three volumes of oxygen, on gets two volumes of SeO2. Here, .90 yield.
Volume SeO2= .90*2/3 * 5.5
check my thinking.
I am a little disturbed that we are calculating SeO2 as a gas. If SiO2 is a solid, the would we not expect SeO2 to be a solid? My Merck Index lists it as a solid. At any rate, the other post is correct but here is how you do it in mols.
2H2Se + 3O2 ==> 2SeO2 + 2H2O(g)
mols oxygen = 5.5L/22.4 = 0.2455.
mols SeO2 produced = 0.2455 mols O2 x (2 mols SeO2/2 mols O2) = 0.164 mols SeO2.
vol SeO2 = 0.164 mol x 22.4 L/mol =3.67 L.
The yield is only 90%; therefore, we would obtain 3.67 x 0.9 = 3.3 L
I hope this helps.
Your thinking in the first part is correct regarding the yield and the stoichiometry of the reaction. Let's break down the calculation:
1. First, we need to balance the chemical equation:
2H2Se + 3O2 → 2SeO2 + 2H2O
2. According to the balanced equation, for every 3 volumes (moles) of O2, we get 2 volumes (moles) of SeO2. Since we know the volume of O2 is 5.50 L, we can calculate the moles of O2:
moles O2 = volume O2 / molar volume of gas at given condition
moles O2 = 5.50 L / 22.4 L/mol (molar volume of an ideal gas at STP)
moles O2 = 0.2455 mol (rounded to 4 decimal places)
3. From the balanced equation, we can see that the stoichiometric ratio between O2 and SeO2 is 3:2. Multiply the moles of O2 by the appropriate ratio to find the moles of SeO2 produced:
moles SeO2 = moles O2 * (2 moles SeO2 / 3 moles O2)
moles SeO2 = 0.2455 mol * (2/3)
moles SeO2 = 0.1637 mol (rounded to 4 decimal places)
4. Finally, to convert the moles of SeO2 to volume, we can use the molar volume of a gas at the given conditions:
volume SeO2 = moles SeO2 * molar volume of gas at given condition
volume SeO2 = 0.1637 mol * 22.4 L/mol
volume SeO2 = 3.67 L (rounded to 2 decimal places)
However, it is worth noting that the yield of the reaction is given as 90%. Therefore, we need to account for this yield in the final answer:
actual volume SeO2 = volume SeO2 * yield
actual volume SeO2 = 3.67 L * 0.90
actual volume SeO2 = 3.3 L (rounded to 1 decimal place)
So, based on the given information, 5.50 L of oxygen would produce approximately 3.3 liters of selenium dioxide, taking into account a yield of 90%.