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

3 Answers

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  • Nip
    Lv 4
    9 years ago
    Favorite Answer

    2H2(g) + O2(g) -----------> 2H20(l)

    So according to stoichiometry

    nH2:nO2=2:1

    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

    VH2:VO2=2:1

    If VH2=8.0L

    Then VO2=8.0/2 L=4.0L

  • 9 years ago

    From the balanced equation:

    2 litres of H2 will react with 1 litre of O2

    Therefore:

    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)

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