in need of some help with this question please~

when 4g of oxygen gas reacts with excess of hydrogen gas, the volume of water vapour produced at (STP) is ....... L ( H = 1 , O = 16).

a) 22.4
b) 44.8
c) 11.2
d) 5.6

an explanation would be great for future reference <3 thank you in advance!

in need of some help with this question pleasewhen 4g of oxygen gas reacts with excess of hydrogen gas the volume of water vapour produced at STP is L H 1 O 16a class=

Respuesta :

Answer:

5.60 Liters at STP

Explanation:

First you need the balanced reaction equation

2 H2 +  O2   ====>   2  H2O

4 g of oxygen ( O2 )  is   4/(15.999*2) moles of O2     (using periodic table)

  Twice the number of moles of H2O re produced   (from equation)

     at STP the volume of a mole of gas is 22.4 liter/ mole

       put all of this together  

         2 *   4 / (15.999*2) moles * 22.4 L/mole = 5.60 L