A. The mass of O₂ needed to burn 36.1 g of B₂H₆ is 123.77 g
B. The number of mole of water, H₂O produced from 19.2 g of B₂H₆ is 1.372 mole
B₂H₆ + 3O₂ —> 2HBO₂ + 2H₂O
Molar mass of B₂H₆ = (2×11) + (1×6) = 28 g/mol
Mass of B₂H₆ from the balanced equation = 1 × 28 = 28 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 3 × 32 = 96 g
SUMMARY
From the balanced equation above,
28 g of B₂H₆ required 96 g of O₂
From the balanced equation above,
28 g of B₂H₆ required 96 g of O₂
Therefore
36.1 g of B₂H₆ will require = (36.1 × 96) / 28 = 123.77 g of O₂
Thus, 123.77 g of O₂ is required for the reaction
We'll begin by calculating the number of mole of in 19.2 g of B₂H₆. This can be obtained as follow:
Mole = mass / molar mass
Mole of B₂H₆ = 19.2 / 28
Mole of B₂H₆ = 0.686 mole
Finally, we shall determine the number of mole of water produced. This is illustrated below:
B₂H₆ + 3O₂ —> 2HBO₂ + 2H₂O
From the balanced equation above
1 mole of B₂H₆ reacted to produce 2 moles of H₂O
Therefore, H₂O
0.686 mole of B₂H₆ will react to produce = 0.686 × 2 = 1.372 mole of H₂O
Thus, 1.372 mole of H₂O was obtained from the reaction
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