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If 1.20 grams of sodium metal are reacted with 3.00 ml of water how many grams of NaOH will be produced?
Cowgirl289
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
Wiki User
∙ 11y ago
Wiki User
∙ 12y ago2Na + 2H O -------->2Na OH +H
2 2
Wiki User
∙ 14y ago2Na + 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
Wiki User
∙ 12y agoThe mass of NaOH is 79,9942 g.
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