Question

How many grams of sodium can react with 750 mL of a 6.0 M solution of sulfuric acid, H2SO4? __ Na + __ H2SO4 __ Na2SO4 + __ H2

Answer 1

Answer:

207 grams of sodium can react with 750 mL of a 6.0 M solution of sulfuric acid.

Explanation:

Moles of sufuric acid :

$6.0 M=\frac{Moles}{0.75 L}$

Moles of sulfuric acid = 4.5 moles

$2Na + H_2SO_4\rightarrow Na_2SO_4 +H_2$

According to reaction , 1 mol of sulfuric acid reacts with 2 mol of sodium

Then 4.5 moles of sulfuric acid will react with:

$4.5 moles\times 2=9.0 mol$ of sodium

Mass of sodium =$9.0 mol\times 23 g/mol=207 g$

207 grams of sodium can react with 750 mL of a 6.0 M solution of sulfuric acid.

Answer 2

Answer : The mass of sodium is, 207 grams

Explanation : Given,

Concentration = 6 M

Volume = 750 ml = 0.750 L

First we have to calculate the moles of sulfuric acid.

$\text{Moles of }H_2SO_4=\text{Concentration}\times \text{Volume}=6mole/L\times 0.750L=4.5mole$

Now we have to calculate the moles of sodium.

The balanced chemical reaction is,

$H_2SO_4+2Na\rightarrow Na_2SO_4+H_2$

From the balanced reaction, we conclude that

As, 1 mole of $H_2SO_4$ react with 2 moles of Na

So, 4.5 mole of $H_2SO_4$ react with $2\times 4.5=9$ moles of Na

Now we have to calculate the mass of sodium.

$\text{Mass of Na}=\text{Moles of Na}\times \text{Molar mass of Na}=9mole\times 23g/mole=207g$

Therefore, the mass of sodium is, 207 grams