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Para la siguiente reacción: 2NH 3(g) + CO 2(g) ———> (NH 2 ) 2 CO (l) + H 2 O (l) a) ¿Cuántos gramos de NH 3 se necesitan para formar 720 g de (NH 2 ) 2 CO? b) ¿Cuántos moléculas de agua se obtienen a partir de 9 litros de CO 2 ayudenme porfa

Sagot :

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Answer:

a. 408g de NH3 son necesarios

b. 2.42x10²³ moléculas de agua se obtienen

Explanation:

a. Basados en la reacción, 2 moles de NH3 producen una mol de (NH2)2CO. Para resolver esta pregunta debemos convertir la masa de (NH2)2CO a moles. Estas moles a moles de NH3 y su respectiva masa:

Moles (NH2)2CO -Masa molar: 60.06g/mol-

720g * (1mol / 60.06g) = 11.99 moles (NH2)2CO

Moles NH3:

11.99 moles (NH2)2CO * (2mol NH3 / 1mol (NH2)2CO) = 23.98 moles NH3

Masa NH3 -17.031g/mol-:

23.98 moles NH3 * (17.031g / mol) = 408g de NH3 son necesarios

b. 1mol de CO2 produce  1mol de agua. Se debe convertir el volumen a moles usando PV = nRT. Estas moles = Moles de agua. 1mol = 6.022x10²³ moléculas:

Asumiendo STP:

PV = nRT; PV / RT = n

Donde P = 1atm a STP

V = 9L

R = 0.082atmL/molK

T = 273.15K a STP

1atm*9L / 0.082atmL/molK*273.15K = n

n = 0.402 moles CO2 = Moles H2O

Moléculas:

0.402 moles agua * (6.022x10²³ moléculas / 1mol) =

2.42x10²³ moléculas de agua se obtienen

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