Both equation I and II are balanced, but equation II is the correct way to write the balanced equation. Why do you think this is? (Hint: Think fractions.)
I) 2 SnO2 + 4 H2 → 2 Sn + 4 H2O
II) SnO2 + 2 H2 → Sn + 2 H2O

Because you have to give the simplest ratio of equation. Equation I can be simplified by dividing all the coefficients by 2 to make it into equation II.

*Every equation has to be in the simplest form of coefficients.

Ratio of Equation I :

2 : 4 : 2 : 4

Ratio of Equation II :

1 : 2 : 1 : 2

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g