Application 2: we mixed 400 g of carbon dioxide CO2 with water H2O, oxygen and glucose C6H12O6 are formed according to the following chemical equation: CO2 (g) + H2O C6H12O6 + O2 (g)
1- Balance the above reaction.
2- Calculate the number of mole of water H2O needed, and that of products.
3- Deduce the mass of C6H12O6 and the volume of oxygen O2 released.
Given: molar mass in g. mol-1: C: 12 H: 1 O: 16

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

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