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For the following reaction, how many grams of H2O will be produced if you react 3.7 grams of B2H6?B2H6 + 3 O2 → 2 HBO2 + 2 H2O

Sagot :

[tex]B_2H_6+3O_2\rightarrow2HBO_2+2H_2O[/tex]

From the chemical equation we see that 1 mole of diborane (B2H6) will produce 2 moles of water.

We will firstly calculate the moles of diborane:

[tex]\begin{gathered} _nB_2H_6=\frac{mass}{molar\text{ }mass} \\ \\ _nB_2H_6=\frac{3.7g}{27.7\text{ }gmol^{-1}} \\ _nB_2H_6=0.134\text{ }moles \end{gathered}[/tex]

One mole of diborane produces 2 moles of water then to find the moles of water that is produced we:

[tex]\begin{gathered} _nH_2O=2\times0.134\text{ }mole \\ _nH_2O=0.268 \end{gathered}[/tex]

We then convert the moles of the water to mass:

[tex]\begin{gathered} mass=moles\times molar\text{ }mass\text{ }H_2O_ \\ \\ mass=0.268\text{ }mol\times18\text{ }gmol^{-1} \\ \\ mass=4.824g \end{gathered}[/tex]

Mass of water is 4.824g