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mrs_skeptik [129]
3 years ago
7

How many molecules of o2 are required to form 13.5 g of water vapor?

Chemistry
1 answer:
Viefleur [7K]3 years ago
3 0
13.5g H2O = 13.5/(2+16) = 0.75 mol H2O

Since for every O2 you can form 2 H2O, 0.75/2 = 0.375 mol of O2 will be needed to form 0.75 mol of H2O. 

Converting from moles to molecules:

0.375 mol * 6.022 * 10^23 = 2.26 * 10 ^23 molecules
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inessss [21]

Answer: 2SO_3(g)+2NO(g)\rightleftharpoons 2SO_2(g)+2NO_2(g)

Explanation:

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

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K_p=K_c when \Delta n_g = 0

a) 4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)

\Delta n_g = change in number of moles of gas particles = 8-7=1

b) 2SO_3(g)+2NO(g)\rightleftharpoons 2SO_2(g)+2NO_2(g)

\Delta n_g = change in number of moles of gas particles = 4-4=0

c) 4N_2(g)+2O_2(g)\rightleftharpoons 4N_2O(g)

\Delta n_g = change in number of moles of gas particles = 4-6=-2

d) 6SO_2(g)+3O_2(g)\rightleftharpoons 6SO_3(g)

\Delta n_g = change in number of moles of gas particles = 6-9=-3

Thus for reaction ,2SO_3(g)+2NO(g)\rightleftharpoons 2SO_2(g)+2NO_2(g) , K_p=K_c as \Delta n_g = 0

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