Question

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

Answer 1

Answer on Question #41279, Chemistry, Other

Task:

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

Answer:

The chemical reaction for this process is:

$$2 \mathrm{Na} + 2 \mathrm{H_2O} = 2 \mathrm{NaOH} + \mathrm{H_2} \uparrow$$

The amount of sodium in the process is:

$$v(\mathrm{Na}) = \frac{m}{A_r(\mathrm{Na})}$$

where m = mass, grams;

$A_r$ = atomic mass, gram/mol.

$$v(\mathrm{Na}) = \frac{299}{23} = 13 \text{ moles}$$

We can assume from the reaction, that:

$$\frac{v(\mathrm{Na})}{v(H_2)} = \frac{2}{1}$$

That is why:

$$v(H_2) = \frac{13}{2} = 6.5 \text{ moles}$$

For STP conditions:

$$v(H_2) = v_m \cdot v$$ $$V(H_2) = 22.4 \times 6.5 = 145.6 \text{ L}$$

The mass from the amount of substance could be found as:

$$m(H_2) = v(H_2) \cdot A(H_2) = 6.5 \cdot 2 = 13 \text{ g}$$

The number of molecules could be calculated over Avogadro's number ($N_A = 6.02 \cdot 10^{23} \text{ mol}^{-1}$):

$$N = N_A \cdot v(H_2) = 6.02 \cdot 10^{23} \cdot 6.5 = 39.13 \cdot 10^{23}$$

http://www.AssignmentExpert.com/