please help i cant remember how to solve these. its for my chem finals.?

Question

12.
mass of sodium would be produced when 2.3 g of sodium reacts with .56 L of chlorine gas. assume Stp & 100% yield. Calculations and balanced equation required.

49.
The salt produced when magnesium hydroxide reacts with sulfuric acid. Balanced equation, chemical formula, chemical name and common name

50. The amount of NaCl produced from 10.o g of sodium hydrogen carbonate reacting with an excess of hydrochloric acid. assume 75% yield. Balance equation and calculations required

Answer 1

12. The reaction is 2 Na + Cl2 -> 2 NaCl.
(1) Find moles of Cl2 in 0.56 L. 1 mole would occupy 22.4 liters.
(2) From reaction, for every mole of Cl2, 2 g-atoms of Na are needed. Compute the g-moles of Na needed to react with Cl2.
(3) Compute the mass of Na (1 g-mole = 23 g).
It should be LESS than what you have available. If not, sodium is reaction-limiting, and you would have to do plan B.

49. Mg(OH)2 + H2SO4 -> 2 H2O + MgSO4
The salt is magnesium sulfate, an ingredient of "milk of magnesia"

50. The reaction is HCl + NaHCO3 -> NaCl+ CO2
+ H2O. (a) compute moles of sodium bicarb in 10 grams of the material.
(b) Take 75% of that number of moles, the answer will be the moles of NaCl produced.
(c) Compute the mole wt of NaCl. Multiply that by the moles from (b) to get the answer.

Answer 2

I'm not going to get the answers for you but I can tell you what to do:

You need to remember that n (moles)=mass/molar mass.
And at STP, V = n/22.4 (litres).

Good luck.

Answer 3

i'm not sure if this is the answer.u check itok??
49.
Mg(OH)2+H2SO4----------->MgSO4+2H2O
(magnesium sulphate)
50.
NaHCO3+HCl--------->NaCl +H2CO3
1mol 1mol
84gm 58.5gm
84gm yield 58.5gm
so 10gm yield (58.5/84)*10gm
75% of that is 75/100*(585/84)gm
apprx. equal to 5.223
plz check if this is ans if wrong i'm sorry