Chemistry help, please?

Question

How would you go about writing a balanced equation describing the reaction of sodium metal with water to produce hydrogen gas and sodium hydroxide?

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Me + Chemistry = Stress
Also, if anyone is up to it:

Manganese metal can be prepared by the thermite process:
4Al(s) + 3MnO2(s) ---> 3Mn(l) + 2Al2O3(s)

Molar masses being:
4Al(s) = 26.98
3MnO2(s) = 86.94
3Mn(l) = 54.94
2Al2O3(s) = 101.96

a) If you start with 8 moles of Al, how many moles of Mn can be produced? (All other reactants are in excess.)
b)What mass of Al2O3 can be made from 200 g MnO2 and excess Al?

Any help is much appreciated.

Answer 1

How would you go about writing a balanced equation describing the reaction of sodium metal with water to produce hydrogen gas and sodium hydroxide?

2Na(s) + 2H2O(aq) --> H2(g) + 2NaOH(aq)

a) If you start with 8 moles of Al, how many moles of Mn can be produced?

According to the balanced reaction equation, the mole ratio between Al and Mn is 4:3
Thus, 8 moles of Al will produce
(3 x 8)/4 = 6 moles of Mn.

b)What mass of Al2O3 can be made from 200 g MnO2 and excess Al?

4Al(s) + 3MnO2(s) ---> 3Mn(l) + 2Al2O3(s)

m(MnO2) = 200g
n(MnO2) = m(MnO2) / M(MnO2)
n = moles ; m = mass(g) ; M =molar mass(gmol^-1
)
n(MnO2) = 200g / (86.94)gmol^-1
n(MnO2) = 2.300437083 mol


According to the balanced reaction equation, mole ratio between Al2O3 and MnO2 is 2:3

n(Al2O3) = n(MnO2) / (3/2)
n(Al2O3) = 2.300437083 mol / 1.5
n(Al2O3) = 1.533624722 mol


m(Al2O3) = n(Al2O3) x M(Al2O3)
m(Al2O3) = 1.533624722 mol x (101.96)
gmol^-1

m(Al2O3) = 156.3683767g


Thus the answer is 156.37g(2 d.p.) of Al2O3


hope this helps:-)

Answer 2

How would you go about writing a balanced equation describing the reaction of sodium metal with water to produce hydrogen gas and sodium hydroxide?
sodium metal: Na
Water: H2O
Hydrogen gas: H2
sodium hydroxide: NaOH

Na + H2O --> H2 + NaOH
H2O can also be written as HOH, so you can say
Na + HOH --> H2 + NaOH
so you have 1 H on the left, 2 on the right, so fix that
Na + 2HOH --> H2 + NaOH
now you need 2 OH and therefore 2 Na
2Na + 2HOH --> H2 + 2NaOH
there you go!





a) If you start with 8 moles of Al, how many moles of Mn can be produced? (All other reactants are in excess.)
8 moles Al * (3 moles Mn / 4 moles Al) = 6 moles Mn
(moles of Al cancel out)


b)What mass of Al2O3 can be made from 200 g MnO2 and excess Al?
200 g MnO2 * (1 mole MnO2 / 86.94 g MnO2) * (2 mole Al2O3 / 3 moles MnO2) * (101.96 grams Al2O3 / 1 mole Al2O3) = 156.37 grams Al2O3

Answer 3

2Na + 2H2O -> 2 NaOH + H2(g)

Re: Thermite Rxn.
From reaction, for each 4 g-atoms of Al reacted, 3 g-atoms of Mn are formed. So by proportion, 8 Al's produce 6 Mn's.

CATTBARF's GOLDEN RULE: Do your reaction comparisons in moles.

200 g MnO2 /87 g/g-mole = 2.4 mole appx
From rxn, 2 moles of Al2O3 are created from ech 3 moles of MnO2 reacted. By proportion, appx 1.6 mole of Al2O3 is created. So weight= 1.6 mole x 102 g/g-mole= 170 gm appx.

Answer 4

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

a) 8 mol Al x (3 mol Mn / 4 mol Al )= 6 mol Mn
(The (3 mol Mn / 4 mol Al ) being the mole to mole ratio from your reaction.)

b) 200g MnO2 x (1 mol MnO2 / 86.94g MnO2) x (2 mol Al2O3 / 3 mol MnO2) x ( 101.96 g Al2O3 / 1 mol Al2O3)=156.37 g

Answer 5

2Na+2H2O=H2+2NaHO

2 sodium + 2 water = 1 common hydrogen + 2 sodium hydroxide

Answer 6

Na+H2O = Na-OH +!/2 H2 the product is sodium hydroxide and hydrogen.

Answer 7

first find the atomic value below the symbol(s) then, multiply that by the number of components.
ok,ok,ok... the answer to your questionA is 4
and your answer to questionB would be 0 you will need an extra componentadded to AI203 to make MN02.

Answer 8

Very interesting.

Answer 9

Google it