Stoichiometry problem....PLEASE help, i'm so confused-10 points?
i'm so confused when it comes to these stoichiometry problems. can anyone please help with this problem, definite 10 points to most help thanks-
Given: H2SO4 + NaOH ---> Na2SO4 + HOH(unbalanced)
it says to balance the equation first
a) if 20.0 grams of H2SO4 and 20.0 grams of NaOH are reacted, how many grams of Na2SO4 will be produced?
thanks a lot!
- 1 decade agoFavorite Answer
look at it this way: the left-hand side (LHS) of the equation
has 3 H, 1 S, 5 O and 1 Na. The right-hand side (RHS) of the equation has 2 Na, 1 S, 5 O and 2 H.
so the first action you need would be to multiply the NaOH on the LHS by 2 (since you need at least 2 Na).
This would then give you: 4H, 1S, 6O, 2Na. (on LHS)
next you would need to multiply the H2O on the RHS by 2 to get a balance of the number of Os and Hs.
This gives you: 2Na, 1S, 6O, 4H. (RHS)
after this you would notice that LHS = RHS, right?
You then would have the basic equation. Now be wary that the equation is basically saying
1 mole of H2SO4 reacting with 2 moles of NaOH will give you 1 mole of Na2SO4 and 2 moles of H2O.
Moles and grams are not equitable. I leave it up to you to work out the relationship of moles to grams (this should be part of your standard definition training) you will need this before you can calculate how much Na2SO4 will be produced.
- Anonymous1 decade ago
h2so4 + 2naoh gives na2so4 + 2 hoh
each atom must balance on each side, so by using 2 molecules of naoh it gives you 2 of hoh (h2o) and balances the na2 on the right hand side of the equation.
that is ---- 4 hydrogen, 1 sulphur, 6 oxygen, and 2 sodium (na) on each side of the equation ---- easy really,hope it helps.
part 2 ---- as two molecules of naoh is needed for a balanced equation, only half of the h2so4 will be reacted. so you will make 30 grams of na2so4 and have 10 grams of unreacted h2so4. ( i think )
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- vvLv 61 decade ago
One mole of H2SO4 requires 2 moles of NaOH; the balanced equation H2SO4 + 2 NaOH ----> 2 H2O and Na2SO4
20 g of H2SO4 = 20/98.08 = 0.204 moles
20 g of NaOH = 20/40 = 0.50 moles
The reaction has NaOH in excess; the limiting reactant is H2SO4; therefore 0.204 moles of Na2SO4 ( molar mass 142.04) will be formed; 0.204 x 142.04 = 28.98 g formed.