Based of the chemical reaction below, if you have 3.56x10^23 molecules of C3H8, how many liters of CO2 gas will be produced at STP? C3H8+5)2-->3CO2+4H2O

39.65L

Explanation:

First, let us calculate the number of mole present in 3.56x10^23 molecules of C3H8.

1 mole of any substance contains 6.02x10^23 molecules as discovered from Avogadro's hypothesis. This means that 1mole of C3H8 contains 6.02x10^23 molecules.

Now, if 1mole of C3H8 contains 6.02x10^23 molecules, then Xmol of C3H8 will contain 3.56x10^23 molecules i.e

Xmol of C3H8 = 3.56x10^23/6.02x10^23 = 0.59mole

The equation for the reaction is given below:

C3H8 + 5O2 —> 3CO2 + 4H2O

From the equation above,

1 mole of C3H8 produced 3moles of CO2.

Therefore, 0.59mole of C3H8 will produce = 0.59 x 3 = 1.77mole of CO2.

Now let us calculate the volume occupied by 1.77mole of CO2. This is illustrated below:

1 mole of any gas occupy 22.4L at stp.

Therefore, 1.77mole of CO2 will occupy = 1.77 x 22.4 = 39.65L

Therefore, 39.65L of CO2 are produced.

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