If 1.20 grams of sodium metal are reacted with 3.00 ml of water how many grams of NaOH will be produced?

Answer 1

First write balance equation

2Na + 2H2O --> 2NaOH + H2

molar mass Na = 22.99 g/mol

molar mass H2O = 18.02 g/mol

molar mass NaOH = 40.0 g/mol

Determine limiting reagent.

1.20 g Na * 1mol Na/22.99g Na = 0.05219 (0.0522) mol Na

Since 2Na = 2NaOH in balanced equation, The mol of NaOH is also 0.0522

3ml h20 = 3 gram h20 (1ml^3=1g^3)

3g H20 * 1mol/18gH20 = 0.167 (0.17) mol H20

So .17 ml H20 equals .17 mol NaOH

and .0522 mole Na equals .0522 mol NaOH

The smaller one is the limiting reagent, which in this case is Na

0.0522mol NaOH*40g NaOH/1mol NaOH = 2.09 gram NaOH

Answer 2

The chemical reaction between sodium metal and water produces sodium hydroxide and hydrogen gas. Given the reaction:
2 Na + 2 H2O → 2 NaOH + H2.
First, calculate the moles of sodium reacted using its molar mass. Then, use the stoichiometry of the reaction to find the moles of NaOH produced. Finally, convert the moles of NaOH to grams using its molar mass.

Answer 3

2Na + 2H O -------->2Na OH +H

2 2

Answer 4

2Na + 2H2O >> 2NaOH + H2

20 grams Na (1mol Na/22.99g )(2mol NaOH/2mol Na )(39.998g NaOH/1mol NaOH )

= 34.8 grams of NaOH produced

Answer 5

The mass of NaOH is 79,9942 g.