Question

Знайти масу та об'єм водню та кисню, які при взаємодії утворилис 60 г води, умови нормальні​

Answer 1

Ответ:

ответ смотри во вложении

Answer 2

Calculation of the Mass and Volume of Hydrogen and Oxygen

To find the mass and volume of hydrogen and oxygen that react to form 60 g of water, we need to use the balanced chemical equation for the reaction between hydrogen and oxygen to form water.

The balanced chemical equation for the reaction is:

2H₂ + O₂ → 2H₂O

From the balanced equation, we can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

To calculate the mass and volume of hydrogen and oxygen, we need to use the molar masses of hydrogen (H₂) and oxygen (O₂).

The molar mass of hydrogen (H₂) is approximately 2 g/mol, and the molar mass of oxygen (O₂) is approximately 32 g/mol.

Calculation of the Mass of Hydrogen:

Since 2 moles of hydrogen react to form 2 moles of water, we can calculate the mass of hydrogen using the following equation:

Mass of hydrogen = (Molar mass of hydrogen) x (Number of moles of hydrogen)

Number of moles of hydrogen = Number of moles of water

Number of moles of water = Mass of water / Molar mass of water

The molar mass of water (H₂O) is approximately 18 g/mol.

Number of moles of water = 60 g / 18 g/mol = 3.33 mol

Mass of hydrogen = 2 g/mol x 3.33 mol = 6.66 g

Therefore, the mass of hydrogen that reacts to form 60 g of water is approximately 6.66 g.

Calculation of the Mass of Oxygen:

Since 1 mole of oxygen reacts to form 2 moles of water, we can calculate the mass of oxygen using the following equation:

Mass of oxygen = (Molar mass of oxygen) x (Number of moles of oxygen)

Number of moles of oxygen = (1/2) x (Number of moles of water)

Number of moles of oxygen = (1/2) x 3.33 mol = 1.67 mol

Mass of oxygen = 32 g/mol x 1.67 mol = 53.44 g

Therefore, the mass of oxygen that reacts to form 60 g of water is approximately 53.44 g.

Calculation of the Volume of Hydrogen and Oxygen:

To calculate the volume of hydrogen and oxygen, we need to use the ideal gas law equation:

PV = nRT

Where: P = Pressure (assumed to be at normal conditions) V = Volume n = Number of moles R = Ideal gas constant T = Temperature (assumed to be at normal conditions)

Since the conditions are assumed to be at normal conditions, we can use the following values: P = 1 atm R = 0.0821 L·atm/(mol·K) T = 273 K

For hydrogen: Number of moles of hydrogen = 3.33 mol (as calculated earlier)

Using the ideal gas law equation, we can calculate the volume of hydrogen:

V(Hydrogen) = (Number of moles of hydrogen) x (R) x (T) / (P)

V(Hydrogen) = 3.33 mol x 0.0821 L·atm/(mol·K) x 273 K / 1 atm

V(Hydrogen) = 73.44 L

Therefore, the volume of hydrogen that reacts to form 60 g of water is approximately 73.44 L.

For oxygen: Number of moles of oxygen = 1.67 mol (as calculated earlier)

Using the ideal gas law equation, we can calculate the volume of oxygen:

V(Oxygen) = (Number of moles of oxygen) x (R) x (T) / (P)

V(Oxygen) = 1.67 mol x 0.0821 L·atm/(mol·K) x 273 K / 1 atm

V(Oxygen) = 37.64 L

Therefore, the volume of oxygen that reacts to form 60 g of water is approximately 37.64 L.

Summary:

Based on the calculations, the mass of hydrogen that reacts to form 60 g of water is approximately 6.66 g, and the mass of oxygen is approximately 53.44 g. The volume of hydrogen is approximately 73.44 L, and the volume of oxygen is approximately 37.64 L.

Please note that these calculations are approximate and assume ideal conditions.