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3 years ago

What is the theoretical yield of so3 produced by the quantities described in part a? express your answer numerically in grams?

1 answer:

8 0

Sulfur reacts with oxygen to yield SO3 as shown in the equation below;
2S(g)+ 3O2(g) = 2SO3(g)
From part A 7.49 g of S were used.
The atomic mass of sulfur is 32.06 g/mol
Hence, the number of moles of sulfur used
7.49 / 32.06 = 0.2336 moles
The mole ratio of S : SO3 is 1:1
Thus the mass of SO3 will be ( 1 mol of SO3= 80.06 g)
0.2336 moles × 80.06 = 18.7 g

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2 years ago

Consider the reaction below.

Law Incorporation [45]

Answer:

<u>First choice: 0.042</u>

Explanation:

Given decomposition reaction:

1PCl₅ (g) ⇄ 1PCl₃ + 1Cl₂(g)

Equilibrium constant:

$K_{eq}=\frac{[PCl_3]^1[Cl_2]^1}{[PCl_5]^1}$

Stoichiometric coefficients and powers equal to 1 are not usually shown as they are understood, but I included them in order to shwow you how they intervene in the equilibrium expressions: each concentration is raised to a power equal to the respective stoichiometric coefficient in the equilibrium equation.

So, your calculations are:

$K_{eq}=\frac{(0.020M)(0.020M)}{0.0095M}=0.042M$

6 0

4 years ago

Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty

anastassius [24]

Answer :

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression of $HC_2H_3O_2$ will be:

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression of $Co(H_2O)_6^{3+}$ will be:

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression of $CH_3NH_3^+$ will be:

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

3 0

3 years ago