What would be the theoretical yield in grams of carbon dioxide in the reaction shown below if 30 g of C6H12O6 were reacted with an excess of oxygen? C6H12O6 + O2 -> CO2 + H2O

Thus 44 g of will be produced  if 30 g of were reacted with an excess of oxygen

Explanation:

To calculate the moles :

The balanced chemical equation is:  

As is the limiting reagent as it limits the formation of product and is the excess reagent.

According to stoichiometry :

1 mole of produce =  6 moles of

Thus 0.17 moles of will produce=  of

 Mass of

Thus 44 g of will be produced  if 30 g of were reacted with an excess of oxygen.

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

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b