Water At STP (chemistry)?
Water can be formed according to the equation: 2H2 + O2 -> 2H20
if 8.0 L of hydrogen is reacted at STP, exactly how many liters of oxygen at STP would be need to allow complete reaction?
Can you show work so I know how to do it?
I think it's 4.0 L
- NipLv 49 years agoFavorite Answer
2H2(g) + O2(g) -----------> 2H20(l)
So according to stoichiometry
As both gases are in STP
according to Avogadro law, equal volumes of different gasses contains equal number of moles
So mole ratio= volume ratio
Then VO2=8.0/2 L=4.0L
- Trevor HLv 79 years ago
From the balanced equation:
2 litres of H2 will react with 1 litre of O2
8 litres of H2 will react with 4 litres of O2
Just read the balanced equation.
- 4 years ago
first of all, the mole ratio of Mg to HCl is a million:2. From the quantities it somewhat is sparkling that HCl is the restricting reactant, i.e. HCl runs out first (because of the fact with the intention to react with 0.288 of Mg, you like 2x0.288 = 0.576mol of HCl yet you in simple terms have 0.450 mole HCl!), and as quickly because it runs our the reaction stops (because of the fact there will be not greater HCl for the relax Mg (unreacted) to react with!). Now be conscious that for each 2mol of HCl the reaction provides upward push to 1mol of H2 gas in accordance to the stoichiometry of the reaction. So for 0.450 mol HCl (restricting reactant) you ought to get carry of 0.450/2 mol of H2 gas. We additionally comprehend that 1mol of an perfect gas has a volume of twenty-two.4 L at STP. So assuming the perfect habit for the H2(g) generated, its volume is = (0.450/2) x 22.4L (at STP)