Question

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!
​

Answer 1

5.60 Liters at STP

Step-by-step 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