Lamonie
contestada


Sodium metal reacts with water to produce sodium hydroxide and
hydrogen gas. Determine the amount of hydrogen gas produced if 25.6
grams of sodium reacts with excess water.

Respuesta :

Eduard22sly

Answer:

1.11g of H2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2Na + 2H2O —> 2NaOH + H2

Next, we shall determine the mass of Na that reacted and the mass of H2 produced from the balanced equation. This is illustrated below:

Molar mass of Na = 23g/mol

Mass of Na from the balanced equation = 2 x 23 = 46g

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 1 x 2 = 2g

From the balanced equation above, 46g of Na reacted to produce 2g of H2.

Finally, we shall determine the mass of H2 produced by the reaction of 25.6g of Na. This is illustrated below:

From the balanced equation above, 46g of Na reacted to produce 2g of H2.

Therefore, 25.6g of Na will react to produce = (25.6 x 2)/46 = 1.11g of H2.

Therefore, 1.11g of H2 is produced from the reaction.

sebassandin

Answer:

[tex]n_{H_2}=0.557molH_2\\\\m_{H_2}=1.11gH_2[/tex]

Explanation:

Hello,

In this case, such chemical reaction is represented by:

[tex]2Na+2H_2O\rightarrow 2NaOH+H_2[/tex]

Thus, for 25.6 grams of sodium, whose molar mass is 23 g/mol, reacting in excess water, the amount of hydrogen gas, whose molar mass is 2 g/mol, that is obtained is computed by using the 2:1 molar ratio between sodium and hydrogen:

[tex]m_{H_2}=25.6gNa*\frac{1molNa}{23gNa} *\frac{1molH_2}{2molNa} *\frac{2gH_2}{1molH_2} \\\\m_{H_2}=1.11gH_2[/tex]

Or in moles:

[tex]n_{H_2}=25.6gNa*\frac{1molNa}{23gNa} *\frac{1molH_2}{2molNa} \\\\n_{H_2}=0.557molH_2[/tex]

Regards.

Otras preguntas