For each of these equations, determine the change in the number of moles of gas, δngas.?
A.) (NH4)2 CO3(s) = 2NH3(g) + CO2(g) + H2O(g)

B.) H2(g) + Cl2(g) = 2HCl(g)

C.) 2H2(g) + O2(g) = 2H2O(l)

D.) 2Na(s) + Cl2(g) = 2NaCl(s)

Respuesta :

A.) reactants - 0 moles of gas, products - 4 moles of gas, Δn = 4 moles
B.) reactants - 2 moles of gas, products - 2 moles of gas, Δn = 0 moles
C.) reactants - 3 moles of gas, products - 0 moles of gas, Δn = -3 moles
D.) reactants - 1 mole of gas, products - 0 moles of gas, Δn = -1 moles

A chemical reaction occurs when one or even more molecules is converted into one or so other chemicals, and the further calculation can be defined as follows:

Using formula:

[tex]\bold{\Delta n = \text{number of moles of the gaseous products - number of moles of gaseous reactants}}[/tex]Following are the solution to the given points:

A)

[tex](NH_{4})_{2}\ CO_{3}\ (s) \longrightarrow 2NH_{3}\ (g) + CO_{2}\ (g) + H_{2}O\ (g)\\\\ \Delta n=4-0\ =4\ moles[/tex]

B)

[tex]H_2\ (g) + Cl_2\ (g) \longrightarrow 2HCl\ (g)\\\\ \Delta n= 2-2 =0\ moles[/tex]

C)

[tex]2H_2\ (g) + O_2\ (g) \longrightarrow 2H_2O\ (l)\\\\ \Delta n= 0-3=-3 \ moles[/tex]

D)

[tex]2Na\ (s) + Cl_2\ (g) = 2NaCl\ (s)\\\\\Delta n= 0-1=-1\ moles[/tex]

Therefore, the answer is "4,0,-3, and -1".

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