Can anyone answer this Gas Stoichiometry question with work?

Question

1)How many liters of water can be made from 34 grams of oxygen gas and 6.0 grams of hydrogen gas at STP? What is the limiting reactant for this reaction?

Answer 1

molar mass H2 = 2 g/mol ( not 4 as in the other answer)

molar mass O2 = 32 g/mol

2H2 + O2 >> 2H2O

34 / 32 = 1.06 moles O2

6 / 2 = 3 moles H2

the ratio between H2 and O2 is 2 : 1 so O2 is the limiting reactant

we get 2 x 1.06 moles H2O = 2.12

at STP

p= 1 atm

T = 273 K

pV = nRT

V = nRT / p = 2.12 x 0.0821 x 273 / 1 = 47.6 L

Answer 2

a million)For the reaction 2H2(g) + O2(g) --->2 H2O(g), how many liters of water could be produced from 5 L of oxygen gasoline and an excess of hydrogen? for gases, volume ratio = mol ratio volume ratio of H2O:O2 = 2:a million consequently volume of water could be made = 5L x 2 = 10L 2)how many liters of water could be produced from fifty 5 grams of oxygen gasoline and an excess of hydrogen at STP? moles of O2 = 55g/32 = a million.71875mol at stp,vol of O2 =22.4L x a million.71875 = 38.5L vol of H2O produced= 38.5Lx 2 = 77L

Answer 3

2H2 + O2 → 2H20
6/4 > 34/32
x/36 = 34/32
x = 38.25 g
38.25*22.4/18 = 47.6 L vapor, which condenses to 38.25 ml liquid