Question

Calculate the volume ,mass and number of molecules of hydrogen liberated when 299 g of sodium react with excess of water at STP.

Answer 1

Answer on Question #41188-Chemistry-Inorganic Chemistry

Question:

Calculate the volume, mass and number of molecules of hydrogen liberated when $299\mathrm{g}$ of sodium react with excess of water at STP.

Solution:

Reaction equation

$$2 N a + 2 H _ {2} O \rightarrow 2 N a O H + H _ {2} \uparrow$$

Amount of substance of sodium is:

$$\nu (N a) = \frac {m (N a)}{A r (N a)} = \frac {2 9 9 g}{2 2 . 9 8 m o l / g} = 1 3 m o l$$

From reaction equation:

$$\frac {\nu (N a)}{\nu (H _ {2})} = \frac {2}{1}$$

Thus,

$$\nu (H _ {2}) = 1 3 * 2 = 2 6 m o l$$

For STP:

$$V = V _ {m} * \nu$$

Calculation:

$$V (H _ {2}) = 2 2. 4 \frac {L}{m o l} * 2 6 m o l = 5 8 2. 4 L$$

Calculation mass from amount of substance:

$$m (H _ {2}) = \nu (H _ {2}) * M r (H _ {2}) = 2 6 m o l * 2 \frac {g}{m o l} = 5 2 g$$

Calculating number of molecules using Avogadro's number:

$$N = N _ {A} * \nu (H _ {2}) = 6. 0 2 * 1 0 ^ {2 3} m o l ^ {- 1} * 2 6 m o l = 1. 5 6 * 1 0 ^ {2 5}$$