Question

3).The photosynthesis reaction is 6CO2 (g) + 6H2O (g) (Arrow) C6H12O6(s) + 6O2 (g). If the reaction requires 2.82 x 10^3 kJ/ mole of C6H12O6 formed: a.) How many grams of C6H12O6 can be produced from 4.20 x 10^4 kJ? b). How much energy is required to produced 250g of C6H12O6? c.)How many liters (at STP) of O2 are produced with 250g of C6H12O6? ͢͢͢͢͢

4.) Given the equation; N2 (g) + 3H2(g) (arrow) 2NH3(g). If 7.00L of N2 ( at STP) reacts with 15.0 L of H2, A.) How many L of NH3 is formed? B.) How many L of the excess gas remains unreacted? C.) How many grams of NH3 are formed?

2.) Place the following sets of elements in order of increasing size and give rational for answer . [Al, P, Na, Ar], [Ca, I , P, Br].

3.) Lead and Carbon are in the same column and therefore have the same electron configuration ending. Briefly explain why Carbon is a non metal and Lead is a metal.

Answer 1

3) 6CO2(g) + 6H2O(g) ---> C6H12O6(s) + 6O2     dH = 2.82 x 10^3 kJ/mol

a) with 4.20 x 10^4 kJ/mol energy we would produce = 4.2 x 10^4 x 180.16/2.82 x 10^3 = 2683.234 g of glucose

b) Energy required = 250 x 2.82 x 10^3/180.16 = 3.913 x 10^3 kJ/mol

c) L of O2 produced by 25o g of C6H12O6 = 6 x 250 x 22.4/180.16 = 186.50 L

4) N2(g) + 3H2(g) ---> 2NH3(g)

a) moles of N2 = 7/22.4 = 0.3125 mols

moles of H2 = 15/22.4 = 0.67 mols

If all of N2 is consumed we would require = 3 x 0.3125 = 0.9375 mols of H2

If all of H2 is consumed we would require = 0.67/3 = 0.223 mols of N2

Since moles of H2 is less than required for complete consumption of N2, this is the limiting reagent

L of NH3 formed = 0.67 x 2 x 22.4/3 = 10.00 L

b) L of excess N2 remained = (0.3125 - 0.223) x 22.4 = 2.005 L

c) grams of NH3 formed = 0.67 x 17.031 = 11.41 g

2) Order of elements in increasing size

Ar < P < Al < Na

and,

Br < P < I < Ca

As we go down the group, additional shells are added, thus the nuclear attraction between protons in the nucelus and electrons in the valence shell decreases, the size increases. As we go from left to right in the periodic table, numbe of shells remain the same, we are adding protons to nucleus and electrons to valence shell, this increases the internulear attraction between protons and electrons and thus size reduces.

3) As we go down the periodic table the size of elements increases due to lowering of internuclear attraction between the valence electrons and the protons in the nucleus. This makes the valence electrons more ready to loose. The metallic character is ability of an element to loose its electron. So increasing size also increases metallic character of element. Carbon with greatest electronegativity in the group would have the highest ionization energy and lead the lowest. Thus, carbon is a non-metal whereas, lead is a metal