How many grams of water are produced from the reaction of 1L of 2 molarity hydrochloric acid with 1L of 2 Molarity barium hydroxide?
To determine the amount of water produced from the reaction of 1L of 2 molarity hydrochloric acid (HCl) with 1L of 2 molarity barium hydroxide (Ba(OH)₂), we need to first write and balance the chemical equation for the reaction.
The balanced equation for this reaction is:
2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
From the balanced equation, we can see that for every 2 moles of hydrochloric acid, 2 moles of water are formed.
To solve the problem, follow these steps:
Step 1: Convert the volume of acid and base to moles.
Since the volume is given in liters and the concentration is given in molarity (moles per liter), the number of moles can be calculated by multiplying the volume (in liters) by the molarity.
Number of moles of HCl = 2 moles/L * 1 L = 2 moles
Number of moles of Ba(OH)₂ = 2 moles/L * 1 L = 2 moles
Step 2: Determine the limiting reactant.
To find the limiting reactant, compare the moles of HCl and Ba(OH)₂. The reactant with the smaller number of moles is the limiting reactant.
In this case, both HCl and Ba(OH)₂ have the same number of moles, so none of them is limiting.
Step 3: Calculate the moles of water produced.
Since the reaction is 2 moles of HCl to 2 moles of water, the moles of water produced will be equal to the moles of HCl.
Moles of water produced = 2 moles
Step 4: Convert moles of water to grams.
The molar mass of water (H₂O) is approximately 18 g/mol. Multiply the moles of water by the molar mass to obtain the mass of water.
Mass of water produced = 2 moles * 18 g/mol = 36 grams
Therefore, from the reaction of 1L of 2 molarity hydrochloric acid with 1L of 2 molarity barium hydroxide, 36 grams of water are produced.