Answer to Question #114239 in General Chemistry for Daniel

Question

3. 1.995 g of Cu will react with excess nitric acid to produce how many liters of NO (g) @ SATP?

Answer 1

3Cu + 8HNO₃ => 3Cu(NO₃)₂ + 2NO + 4H₂O

m(Cu) = 1.995 g;

M(Cu) = 63.546 g/mol;

n=m/M;

n(Cu) = 1.995/63.546 = 0.03139 moles;

Proportion according to the reaction:

3 — 2

0.03139 moles — x moles

x = 0.0209 moles of NO were formed;

Standard Ambient Temperature and Pressure (SATP):

Standard ambient temperature is 25 °C or 298.15 K;

Standard pressure is 1 atm.

Ideal gas law:

PV=nRT

Molar gas constant

R=0.082058 L×atm/(K×mol)

(1 atm)×V=(0.0209 mol)×(0.082058 L×atm/(K×mol))×298.15 K;

V = 0.512 L

Answer: 0.512 L