001.
The combustion reaction of diborane (B2H6) with oxygen (O2) is:
B2H6(g) + 3 O2(l) -> 2 HBO2(g) + 2 H2O(l)
We are given that 198.4 g of B2H6 is used, and we want to find the mass of liquid oxygen (LOX) required for the reaction.
From the balanced equation, we can see that for every 1 mole of B2H6, 3 moles of O2 are needed.
We can use the molar mass of B2H6 (37.83 g/mol) to convert the mass of B2H6 to moles:
198.4 g / 37.83 g/mol = 5.24 moles B2H6
Since 3 moles of O2 are needed for every 1 mole of B2H6, we can multiply the number of moles of B2H6 by 3 to find the number of moles of O2 required:
5.24 moles B2H6 x 3 moles O2/1 mole B2H6 = 15.72 moles O2
We can use the molar mass of O2 (32.00 g/mol) to convert the number of moles of O2 to mass:
15.72 moles O2 x 32.00 g/mol = 502.24 g
So, the mass of liquid oxygen (LOX) needed to burn 198.4 g of B2H6 is approximately 502.24 g.
002.
The combustion reaction of diborane (B2H6) with oxygen (O2) is:
B2H6(g) + 3 O2(l) -> 2 HBO2(g) + 2 H2O(l)
We are given that 126.1 g of B2H6 is used, and we want to find the mass of HBO2 produced from the combustion.
From the balanced equation, we can see that for every 1 mole of B2H6, 2 moles of HBO2 are produced.
We can use the molar mass of B2H6 (37.83 g/mol) to convert the mass of B2H6 to moles:
126.1 g / 37.83 g/mol = 3.34 moles B2H6
Since 2 moles of HBO2 are produced for every 1 mole of B2H6, we can multiply the number of moles of B2H6 by 2 to find the number of moles of HBO2 produced:
3.34 moles B2H6 x 2 moles HBO2/1 mole B2H6 = 6.68 moles HBO2
We can use the molar mass of HBO2 (48.05 g/mol) to convert the number of moles of HBO2 to mass:
6.68 moles HBO2 x 48.05 g/mol = 319.55 g
So, the mass of HBO2 produced from the combustion of 126.1 g of B2H6 is approximately 319.55 g.