Find the Density at STP B_2H_6

$B_{2} H_{6}$

Step 1

To find the mass of $1$ mole of $B_{2} H_{6}$ look up the atomic mass of each element and multiply it by the number of atoms contained in each element in the molecule.

mass of $1$ mole of $B_{2} H_{6}$ $=$ $2 (mass of B)$ + $6 (mass of H)$

Step 2

Fill in the atomic masses from the periodic table.

mass of $1$ mole of $B_{2} H_{6}$ $=$ $2 (10.81000041)$ + $6 (1.00999999)$

Step 3

Simplify the result.

Simplify each term.

Multiply $2$ by $10.81000041$ .

mass of $1$ mole of $B_{2} H_{6} = 21.62000083 + 6 (1.00999999) g$

Multiply $6$ by $1.00999999$ .

mass of $1$ mole of $B_{2} H_{6} = 21.62000083 + 6.05999994 g$

Add $21.62000083$ and $6.05999994$ .

mass of $1$ mole of $B_{2} H_{6} = 27.68000078 g$

Step 4

One mole of any gas occupies $22.4 L$ at STP (standard temperature and pressure).

$V = 22.4 L$

Step 5

Density is mass over volume.

$\frac{27.68000078 g}{22.4 L}$

Step 6

Factor $27.68000078$ out of $27.68000078 g$ .

$\frac{27.68000078 (g)}{22.4 L}$

Step 7

Factor $22.4$ out of $22.4 L$ .

$\frac{27.68000078 (g)}{22.4 (L)}$

Step 8

Separate fractions.

$\frac{27.68000078}{22.4} \cdot \frac{g}{L}$

Step 9

Divide $27.68000078$ by $22.4$ .

$1.23571432 \frac{g}{L}$

Step 10

Combine $1.23571432$ and $\frac{g}{L}$ .

$\frac{1.23571432 g}{L}$

Find the Density at STP B_2H_6

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