Question

Газ,выделившийся при нагревании оксида ртути(ll) массой 65,1 пропустили над раскаленным углем массой 1,2 г.Определите количество вещества и объем образовавшегося оксида углерода(lV).

Answer 1

2hgo=2hg+o2 N(hgo) = 65/216 =0.3       0.3/2=x/1 x=0.15
o2+c=co2      N(c)=1.2/12=0.1  N(co2)=0.1 V= 22.4*0.1=2.24л

Answer 2

Calculation of the Amount of Substance and Volume of Carbon Monoxide (IV) Oxide

To determine the amount of substance and volume of carbon monoxide (IV) oxide formed when the gas released from heating mercury (II) oxide with a mass of 65.1 g is passed over hot coal with a mass of 1.2 g, we need to use the balanced chemical equation for the reaction.

The balanced equation for the reaction between mercury (II) oxide and carbon monoxide (IV) oxide is:

HgO + CO → Hg + CO2

From the equation, we can see that 1 mole of mercury (II) oxide reacts with 1 mole of carbon monoxide (IV) oxide to produce 1 mole of mercury and 1 mole of carbon dioxide.

To calculate the amount of substance of carbon monoxide (IV) oxide formed, we need to convert the mass of mercury (II) oxide to moles using its molar mass and then use the stoichiometry of the balanced equation.

The molar mass of mercury (II) oxide (HgO) is 216.59 g/mol.

The molar mass of carbon monoxide (IV) oxide (CO) is 28.01 g/mol.

Using the given masses:

Mass of mercury (II) oxide (HgO) = 65.1 g Mass of coal (C) = 1.2 g

To calculate the amount of substance of carbon monoxide (IV) oxide, we can follow these steps:

1. Calculate the amount of substance of mercury (II) oxide (HgO) using its molar mass: Amount of substance of HgO = Mass of HgO / Molar mass of HgO

2. Use the stoichiometry of the balanced equation to determine the amount of substance of carbon monoxide (IV) oxide (CO): Amount of substance of CO = Amount of substance of HgO

3. Calculate the volume of carbon monoxide (IV) oxide using the ideal gas law: Volume of CO = Amount of substance of CO * Molar volume of gas at standard temperature and pressure (STP)

The molar volume of gas at STP is approximately 22.4 L/mol.

Let's calculate the amount of substance and volume of carbon monoxide (IV) oxide using the given masses:

1. Calculate the amount of substance of mercury (II) oxide (HgO): Amount of substance of HgO = 65.1 g / 216.59 g/mol

2. Use the stoichiometry of the balanced equation to determine the amount of substance of carbon monoxide (IV) oxide (CO): Amount of substance of CO = Amount of substance of HgO

3. Calculate the volume of carbon monoxide (IV) oxide: Volume of CO = Amount of substance of CO * Molar volume of gas at STP

Let's perform the calculations:

1. Calculate the amount of substance of mercury (II) oxide (HgO): Amount of substance of HgO = 65.1 g / 216.59 g/mol = 0.3006 mol

2. Use the stoichiometry of the balanced equation to determine the amount of substance of carbon monoxide (IV) oxide (CO): Amount of substance of CO = Amount of substance of HgO = 0.3006 mol

3. Calculate the volume of carbon monoxide (IV) oxide: Volume of CO = Amount of substance of CO * Molar volume of gas at STP Volume of CO = 0.3006 mol * 22.4 L/mol = 6.73 L

Therefore, the amount of substance of carbon monoxide (IV) oxide formed is 0.3006 mol, and the volume of carbon monoxide (IV) oxide formed is 6.73 L.

Please note that the calculations assume ideal gas behavior and standard temperature and pressure (STP).