How Much Sodium Is Required to Produce 8g of Sodium Hydroxide?

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Sodium reacts with water to produce sodium hydroxide and hydrogen gas, as represented by the balanced equation 2 Na + 2 H2O → 2 NaOH + H2. To prepare 8 grams of sodium hydroxide, the calculation shows that 4.6 grams of sodium are required, based on the molar mass relationship where 23 grams of sodium yield 40 grams of sodium hydroxide. The stoichiometric ratio indicates that 1 mole of sodium produces 1 mole of sodium hydroxide. The discussion confirms the calculation and emphasizes the importance of understanding the mole ratios in chemical reactions. The final answer of 4.6 grams of sodium needed for 8 grams of sodium hydroxide is verified as correct.
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Hello!
Can somebody help to figure out this pls?


Sodium reacts in water to form sodium hydroxide and hydrogen gas. Write a balanced equation for this reaction and calculate how many grams of it are needed to prepare 8 g of sodium hydroxide by this method.
Ar(H)=1 Ar(O)=16 Ar(Na)=23

Balanced equation:
2 Na + 2 H2O --- 2 NaOH + H2

Sodium hydroxide mass: 40 g
I could easily figure out the balanced equation, but I don't know how to do the rest.
Thx!
 
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ignacioserra said:
Hello!
Can somebody help to figure out this please?


Sodium reacts in water to form sodium hydroxide and hydrogen gas. Write a balanced equation for this reaction and calculate how many grams of it are needed to prepare 8 g of sodium hydroxide by this method.
Ar(H)=1 Ar(O)=16 Ar(Na)=23

Balanced equation:
2 Na + 2 H2O --- 2 NaOH + H2

Sodium hydroxide mass: 40 g
I could easily figure out the balanced equation, but I don't know how to do the rest.
Thanks!

Hint :What do you get from equation ? Can you express mass of NaOH and Na in gram molecular mass in the equation ?

From equation what can you say ? How many mass of Na can prepare how many mass of NaOH in gram molecular mass ? Apply unitary method...
 


ignacioserra said:
2 Na + 2 H2O --- 2 NaOH + H2

It tells you: two moles of Na react with two moles of water yielding two moles of NaOH and one mole of hydrogen gas.

Ratio of the reactants and products always stays the same (assuming there is enough other reactants present). In this case ratio of Na to NaOH is 1:1 (which is equivalent to 2:2). 2 moles of Na produce 2 moles of NaOH. 1 mole of Na produces 1 mole of NaOH. 10 moles of Na produce 10 moles of Na. 0.0000000001 mole of Na produces 0.0000000001 mole of NaOH.

It also works the other way. If 2 moles of NaOH were produced, there were 2 moles of Na. If 0.4 moles of NaOH were produced, there were 0.4 moles of Na present.

Now all you need it to convert given mass to moles, convert 8 grams of NaOH to moles, find out how many moles of Na were needed, and convert it to Na mass.
 


So:
NaOH= 16+23+1= 40
Na= 23

23g of Na produce 40 g of NaOH. Right?

So,

23 --- 40
x --- 8 ?

40/5 = 8
23/5= 4.6 g

is that the answer?
 
ignacioserra said:
So:
NaOH= 16+23+1= 40
Na= 23

23g of Na produce 40 g of NaOH. Right?

So,

23 --- 40
x --- 8 ?

40/5 = 8
23/5= 4.6 g

is that the answer?

Yes , that's the answer ! You got the correct answer..:smile:
Well done !

Note : Here 2 moles of NaOH are produced from 2 moles of Na. Ultimately 2 and 2 on both sides canceled , so your answer remained unaffected.

For future reference :
http://www.sciencebugz.com/chemistry/chsolstoich.htm
http://www.sciencegeek.net/Chemistry/taters/Unit4Stoichiometry.htm
http://www.kentchemistry.com/links/Math/LImiting.htm
http://wiki.answers.com/Q/How_do_you_solve_stoichiometry_problems
 
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