Respuesta :
Answer:
The answer to your question is
a) 0.67 moles of O₂
b) 1.34 moles of H₂O
c) 24.12 g of water
d) 10.72 g of O₂
Explanation:
Chemical Reaction
2H₂O₂ (aq) ⇒ 2H₂O (l) + O₂ (g)
Data
1.34 moles of H₂O₂
To solve this problem use proportions
a)
2 moles of H₂O₂ ----------------- 1 mol of O₂
1.34 moles of H₂O₂ -------------- x
x = (1.34 x 1) / 2
x = 0.67 moles of O₂
b)
2 moles of H₂O₂ ------------------ 2 moles of H₂O
1.34 moles of H₂O₂ -------------- x
x = (1.34 x 2)/ 2
x = 1.34 moles of H₂O
c)
18 g of H₂O ------------------- 1 mol
x -------------------- 1.34 moles
x = (1.34 x 18) / 1
x= 24.12 g of water
d)
16 g of O₂ -------------------- 1 mol
x ------------------- 0.67 moles
x = (0.67 x 16) / 1
x = 10.72 g of O₂
Answer:
a) 0.67 moles of O2
b) 1.34 moles H2O
c) 24.15 grams of H2O
d) 21.44 grams O2
Explanation:
Step 1: Data given
Dihydrogen dioxide = H2O2
oxygen gas = O2
Moles H2O2 = 1.34 moles
Molar mass of H2O2 = 34.01 g/mol
Step 2: The balanced equation
2H2O2 → 2H2O + O2
Step 3: Calculate moles of H2O formed
For 2 moles H2O2 we'll have 2 moles H2O produced
For 1.34 moles H2O2 we have 1.34 moles H2O produced
This is 1.34 moles * 18.02 g/mol = 24.15 grams of H2O
Step 4: Calculate moles of O2 formed
For 2 moles H2O2 we'll have 1 mol O2 produced
For 1.34 moles H2O2 we'll have 1.34 /2 = 0.67 moles of O2 produced
This is 0.67 moles * 32.00 g/mol = 21.44 grams O2