Question

Calculati numarul de moli din :  b) 4,48dm la 3 (c.n) N2 ; 4,48L (c.n) NH3 c) 9,033.10 la 23 molecule N2 ; 9,033.10la 26 molecule NH3.Calculati masa a 67,2 dm cubi (c.n) N2 si 67,2 dm cubi (c.n) NH3

Answer 1

Numărul de moli din:

a). V = 4.48 dm³ N₂

$n=\frac{V}{V_M}=\frac{V}{22.4\ dm^3/mol}=\frac{4.48\ {\not}dm^3}{22.4\  {\not}dm^3/mol}=0.2\ mol\ N_2$

b). V = 4.48 L NH₃

$n=\frac{V}{V_M}=\frac{V}{22.4\ L/mol}=\frac{4.48\ {\not}L}{22.4\ {\not}L/mol}=0.2\ mol\ NH_3$

c). 9.033·10²³ molecule N₂

* NA - numărul lui Avogadro (NA = 6.022·10²³ particule (ioni, atomi, molecule)

1 mol .... 6.022·10²³ molecule

x mol .... 9.033·10²³ molecule        ⇒ x = 1.5 mol N₂

d). 9.033·10²⁶ molecule NH₃

1 mol ..... 6.022·10²³ molecule

y mol ..... 9.033·10²⁶                        ⇒ x = 1.5 · 10³ mol NH₃

Masa a 67.2 dm³ de:

a). N₂

$n=\frac{V}{V_M}=\frac{67.2\ {\not}dm^3}{22.4\ {\not}dm^3/mol}=3\ mol\\n=\frac{m}{\mu}\implies m=n\ \cdot\ \mu=3\ {\not}mol \cdot\ 28\ g/{\not}mol = 84\ g\ N_2$

SAU

$67.2\ dm^3=0.0672\ m^3\\\rho_{N_2}=1.25\ kg/m^3\\\rho=\frac{m}{V}\implies m=\rho\ \cdot\ V=1.25\ kg/{\not}m^3\ \cdot 0.0672\ {\not}m^3 =0.084\ kg = 84\ g\ N_2$

b). NH₃

$n=\frac{V}{V_M}=\frac{67.2\ {\not}dm^3}{22.4\ {\not}dm^3/mol}=3\ mol\\n=\frac{m}{\mu}\implies m=n\ \cdot\ \mu=3\ {\not}mol \cdot\ 17\ g/{\not}mol = 51\ g\ NH_3$

SAU

$67.2\ dm^3=0.0672\ m^3\\\rho_{NH_3}=0.72\ kg/m^3\\\rho=\frac{m}{V}\implies m=\rho\ \cdot\ V=0.72\ kg/{\not}m^3\ \cdot 0.0672\ {\not}m^3 =0.048\ kg \simeq 48\ g\ NH_3$