C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
<span>(a) what is the average volume (in cubic meters) required for each iron atom
For this case, the density of Iron would be </span>7.87g/cm³
<span>
V = 9.27 x 10^-26 kg / </span>7.87g/cm<span>³ ( 1 kg / 1000 g)
</span>V = 1.18 x 10-23 cm³<span>
(b) what is the distance (in meters) between the centers of adjacent atoms?
We assume the atoms as cube, so we use the volume of the cube to calculate the distance of the atoms.
V = </span>1.18 x 10-23 cm<span>³ = s</span>³
s = 2.28 x 10^-8 cm
That is correct…….. i think
