Answer:
The answer to your question is: 6.8 g of water
Explanation:
Data
2.6 moles of HCl
1.4 moles of Ca(OH)2
2HCl + Ca(OH)2 → 2H2O + CaCl2
MW 2(36.5) 74 36 g 111 g
73g
1 mol of HCl ---------------- 36.5 g
2.6 mol -------------- x
x = (2.6 x 36.5) / 1 = 94.9 g
1 mol of Ca(OH)2 -------------- 74 g
1.4 mol --------------- x
x = (1.4 x 74) / 1 = 103.6 g
Grams of water
73 g of HCl ------------------ 36g of H2O
94.9 g ------------------- x
x = (94.9 x 36) / 73 = 46.8 g of water
The amount of water is theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH)₂ - C) 46.8.
Given:
2HCl + Ca(OH)₂ -------------> CaCl₂ + 2H2O
in this reaction, 2 moles HCl reacts with 1 mole Ca(OH)₂
which means 2.6 moles of HCl reacts with 2.6 x = 1.3
but we have 1.4 moles Ca(OH)₂ it is the excess reagent
Solution:
HCl is a limiting reagent in this reaction as it allows producing the mole of water on the base of the number of moles it has,
=> 2 mole HCl - 2 moles of H₂O
Here, 2.6 moles of HCl would give 2.6 moles of H₂O only
=> convert moles to mass
moles = mass / molar mass
and,
mass = [tex]\frac{mass}{molar\ mass}[/tex]
mass = 2.6 x 18
= 46.8 g
Thus, the amount of water is theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH)₂ - C) 46.8.
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