Question

суміш що містить 12 г водню та 160 г кисню, привели до умов реакції. Визначте: а) яка з вихідних речовин в надлишку; б) массу речовини, що не вступила в реакцію; в) кількість речовини води, що утворилась; г) масса води, що утворилась

Answer 1

Відповідь:

Пояснення:

n(H2) = m(H2)/ Mr(H2) = 12 г / 2г/моль = 6 моль

n(O2) = m(O2) / Mr(O2) = 160 г / 32 г/моль =  5 моль

Складаємо рівняння реакції

2H2 + O2 = 2H2O

мольне співвідношення за реакцією

2    :     1    :   2

за задачею

6   :      5  , 2 моль кисню - в надлишку, це m (O2) = n(O2) * Mr(O2)=

2 моль * 32 г/ моль = 64 г О2 не вступила в реакцію

За задачею утворилось 6 моль (Н2О)

m(H2O) = n(H2O) * Mr(H2O) =  6 моль * 18 г/ моль = 108 г води.

Answer 2

Given Information:

We have a mixture that contains 12 g of hydrogen and 160 g of oxygen. The mixture is brought to reaction conditions.

Task:

We need to determine: a) The excess reactant. b) The mass of the substance that did not react. c) The amount of water formed. d) The mass of water formed.

Solution:

To solve this problem, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.

To find the limiting reactant, we can compare the number of moles of each reactant. The reactant with the smaller number of moles is the limiting reactant.

Calculating Moles of Hydrogen and Oxygen:

To calculate the number of moles of a substance, we can use the formula:

moles = mass / molar mass

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

a) Moles of Hydrogen: - Mass of hydrogen = 12 g - Molar mass of hydrogen = 2 g/mol - Moles of hydrogen = 12 g / 2 g/mol = 6 mol

b) Moles of Oxygen: - Mass of oxygen = 160 g - Molar mass of oxygen = 32 g/mol - Moles of oxygen = 160 g / 32 g/mol = 5 mol

Determining the Limiting Reactant:

Comparing the number of moles, we can see that there are fewer moles of oxygen (5 mol) compared to hydrogen (6 mol). Therefore, oxygen is the limiting reactant.

Calculating the Excess Reactant:

To determine the excess reactant, we need to calculate the moles of the excess reactant and subtract the moles of the limiting reactant from it.

c) Moles of Excess Hydrogen: - Moles of hydrogen = 6 mol (from previous calculation) - Moles of excess hydrogen = Moles of hydrogen - Moles of oxygen - Moles of excess hydrogen = 6 mol - 5 mol = 1 mol

d) Mass of Excess Hydrogen: - Molar mass of hydrogen = 2 g/mol - Mass of excess hydrogen = Moles of excess hydrogen * Molar mass of hydrogen - Mass of excess hydrogen = 1 mol * 2 g/mol = 2 g

Calculating the Amount of Water Formed:

The balanced chemical equation for the reaction between hydrogen and oxygen to form water is:

2H₂ + O₂ → 2H₂O

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

e) Moles of Water Formed: - Moles of oxygen = 5 mol (from previous calculation) - Moles of water formed = Moles of oxygen * (2 moles of water / 1 mole of oxygen) - Moles of water formed = 5 mol * (2 mol/1 mol) = 10 mol

Calculating the Mass of Water Formed:

To calculate the mass of water formed, we can use the formula:

mass = moles * molar mass

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

f) Mass of Water Formed: - Moles of water formed = 10 mol (from previous calculation) - Molar mass of water = 18 g/mol - Mass of water formed = Moles of water formed * Molar mass of water - Mass of water formed = 10 mol * 18 g/mol = 180 g

Summary of Results:

a) The excess reactant is hydrogen. b) The mass of the substance that did not react is 2 g of hydrogen. c) The amount of water formed is 10 moles. d) The mass of water formed is 180 g.

Please note that the calculations are based on the given information and assumptions about the reaction conditions.