Question

На смесь массой 20 г, содержащую цинк и оксид цинка, подействовали очень разбавленной азотной кислотой. При этом выделился газ, объёмом 1,12 л (н.у). Какова массовая доля оксида цинка в смеси?

Answer 1

ZnO c азотной кислотой не реагируєт. поєтому есть только 1 реакция Zn +HONO2 - Zn(NO3)2+H2. поетому мы найдём нью Н2 = 0.05 по реакции виде что нью Н2=нью Цинка = 0.05 найдём массу цинка смеси Мr=0.05*65=3.25грам. теперь 20-3.25= 16.75 грам оксида цинка.

Answer 2

Calculation of the Mass Fraction of Zinc Oxide in the Mixture

To determine the mass fraction of zinc oxide in the mixture, we need to use the information provided. The mass of the mixture is given as 20 g, and it contains both zinc and zinc oxide. When the mixture is reacted with diluted nitric acid, a gas is released with a volume of 1.12 L at standard temperature and pressure (n.u).

To calculate the mass fraction of zinc oxide, we can use the ideal gas law and the stoichiometry of the reaction. The balanced chemical equation for the reaction between zinc oxide and nitric acid is:

ZnO + 2HNO3 → Zn(NO3)2 + H2O

From the balanced equation, we can see that 1 mole of zinc oxide reacts with 2 moles of nitric acid to produce 1 mole of zinc nitrate and 1 mole of water.

To calculate the moles of gas produced, we can use the ideal gas law equation:

PV = nRT

Where: - P is the pressure (at standard temperature and pressure, P = 1 atm) - V is the volume of the gas (1.12 L) - n is the number of moles of gas (to be determined) - R is the ideal gas constant (0.0821 L·atm/(mol·K)) - T is the temperature (standard temperature and pressure, T = 273.15 K)

Rearranging the equation to solve for n, we have:

n = PV / RT

Substituting the given values, we get:

n = (1 atm) * (1.12 L) / (0.0821 L·atm/(mol·K) * 273.15 K)

Simplifying the equation, we find that n is approximately 0.048 moles.

Since 1 mole of zinc oxide reacts with 1 mole of zinc nitrate, the number of moles of zinc oxide in the mixture is also 0.048 moles.

To calculate the mass of zinc oxide, we can use the molar mass of zinc oxide, which is 81.38 g/mol.

Mass of zinc oxide = moles of zinc oxide * molar mass of zinc oxide = 0.048 moles * 81.38 g/mol

The mass of zinc oxide in the mixture is approximately 3.91 g.

Finally, to calculate the mass fraction of zinc oxide, we divide the mass of zinc oxide by the mass of the mixture and multiply by 100:

Mass fraction of zinc oxide = (mass of zinc oxide / mass of mixture) * 100 = (3.91 g / 20 g) * 100

The mass fraction of zinc oxide in the mixture is approximately 19.55%.

Please note that the calculations provided are based on the information given and assumptions made.