Balanced chemical equation is:
2 H2 + O2 ---> 2 H2O
2 L of H2 reacts with 1 L of O2
for 7.01 L of H2, 3.505 L of O2 is required
But we have 4.11 L of O2
so, H2 is limiting reagent
we will use H2 in further calculation
According to balanced equation
Volume of H2O formed = (2/2)* volume of H2
= (2/2)*7.01
= 7.01 L
Given:
P = 1.0 atm
V = 7.01 L
T = 273.0 K
find number of moles using:
P * V = n*R*T
1 atm * 7.01 L = n * 0.08206 atm.L/mol.K * 273 K
n = 0.3129 mol
Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol
use:
mass of H2O,
m = number of mol * molar mass
= 0.3129 mol * 18.02 g/mol
= 5.637 g
Answer: 5.64 g
QUESTION 4 Consider the reaction between hydrogen gas and oxygen gas to form water 2 H2(g)...
pls show steps 536 QUESTION 4 Consider the reaction between hydrogen gas and oxygen gas to form water 2 H2(g) + O2(g) + 2 H2O(g) How many grams of water could be produced by the reaction of 6 67 liters of hydrogen with 4 11 liters of oxygen at STP? (Hint check for limiting reagent) Click Save and Submit to see and submit. Click Save All Answers to see all answers - F
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