Question

Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calculate the moles in 900. g of hydrogen gas? Set up the math. But DONT DO ANY OF IT. Just leave your answer as a math expression.

Answer 1

Answer:

$n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}$

Explanation:

Assuming that all caculations are at normal pressure and -1.72°C :

$n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}$

Where

$n$ is the number of moles of hydrogen

$n$ is the mass of hydrogen

$\rho$ is the density of hydrogen

Answer 2

Answer:

n = $\frac{900}{0.090*22.4}$

Explanation:

If 1 mole of hydrogen gas occupies 22.4 L, then, the number of moles of hydrogen gas (n) can be found by the volume (V):

1 mole ----- 22.4 L

n ------ V

By a simple direct three rule:

22.4n = V

n = V/22.4

But the volume is mass(m) divided by the density (d)

V = m/d

So

$n = \frac{m}{22.4*d}$

For a mass of 900 g and a density of 0.090 g/L:

n = $\frac{900}{0.090*22.4}$