Calculate the mass of each product formed when 43.82g of diborane (B2H6) reacts with excess water:
B2H6(g) + H2O(l)--> H3BO3(s) + H2(g)
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To calculate the mass of each product formed, we need to determine the stoichiometry of the reaction. The balanced chemical equation gives us the molar ratios between the reactant (diborane) and the products (boric acid and hydrogen gas).
The balanced chemical equation is:
B2H6(g) + H2O(l) --> H3BO3(s) + H2(g)
From the equation, we can see that one mole of diborane (B2H6) reacts with six moles of water (H2O) to produce one mole of boric acid (H3BO3) and three moles of hydrogen gas (H2).
Step 1: Convert the given mass of diborane (B2H6) to moles.
To do this, we need to know the molar mass of B2H6, which is 27.67 g/mol.
Using the formula: Moles = Mass / Molar mass:
Moles of B2H6 = 43.82 g / 27.67 g/mol ≈ 1.583 mol
Step 2: Determine the molar ratios between the reactant and the products.
From the balanced equation, we know that:
1 mole of B2H6 reacts to produce 1 mole of H3BO3
1 mole of B2H6 reacts to produce 3 moles of H2
Step 3: Calculate the mass of each product formed.
Mass of H3BO3 = Moles of B2H6 × Molar mass of H3BO3
Mass of H2 = Moles of B2H6 × Molar mass of H2
The molar masses are:
Molar mass of H3BO3 = 61.83 g/mol
Molar mass of H2 = 2.02 g/mol
Mass of H3BO3 = 1.583 mol × 61.83 g/mol ≈ 97.90 g
Mass of H2 = 1.583 mol × 2.02 g/mol ≈ 3.19 g
Therefore, the mass of boric acid formed is approximately 97.90 g, and the mass of hydrogen gas formed is approximately 3.19 g.