Question

Сколько литров диоксида азота (NO) образуется при растворении 0,59г металлического никеля в разбавленной азотной кислотеNi+HNO3 = Ni(NO3)2+NO+H2O

Answer 1

3Ni + 8HNO3 = 3Ni(NO3)2 + 2NO + 4H2O

n(Ni) = m(Ni)/M(Ni) = 0,59/59 = 0,01 моль

По уравнению видно, что n(NO) = (n(Ni)/3)*2 = (0,01/3)*2 = 0,007 моль

V(NO) = n(NO)*Vm  (Vm - молярный объем, равен 22,4 л/моль при н.у.)

V(NO) = 0,007*22,4 = 0,1568 л.

Answer 2

Calculation of Nitrogen Dioxide (NO) Formation

To calculate the amount of nitrogen dioxide (NO) formed when 0.59g of metallic nickel is dissolved in dilute nitric acid (Ni + HNO3 = Ni(NO3)2 + NO + H2O), we can use stoichiometry and molar mass.

The balanced chemical equation for the reaction is: Ni + 4HNO3 → Ni(NO3)2 + 2NO + 2H2O

First, we need to find the moles of nickel (Ni) using its molar mass. The molar mass of nickel (Ni) is 58.69 g/mol.

The number of moles of nickel (Ni) can be calculated as: moles of Ni = mass of Ni / molar mass of Ni moles of Ni = 0.59g / 58.69 g/mol moles of Ni ≈ 0.01 mol

According to the balanced chemical equation, 1 mole of nickel (Ni) reacts to produce 2 moles of nitrogen dioxide (NO).

Therefore, the moles of nitrogen dioxide (NO) produced will be: moles of NO = 2 * moles of Ni moles of NO = 2 * 0.01 mol moles of NO = 0.02 mol

Now, we can convert the moles of nitrogen dioxide (NO) to liters using the ideal gas law at standard temperature and pressure (STP), where 1 mole of any gas occupies 22.4 liters at STP.

The volume of nitrogen dioxide (NO) at STP will be: volume of NO = moles of NO * 22.4 L/mol volume of NO = 0.02 mol * 22.4 L/mol volume of NO = 0.448 L

Therefore, approximately 0.448 liters of nitrogen dioxide (NO) are formed when 0.59g of metallic nickel is dissolved in dilute nitric acid.