Question

Помогите пожалуйста , дам 30 б 1. Рассчитайте массу сульфата натрия, образовавшегося при нейтрализации гидроксида натрия массой 12 г. серной кислотой. 2NaOH + H2SO4 = Na2SO4 + 2H2O 2. Какова масса кальция, вступившего в реакцию с водой, если при этом образуется гидроксид массой 4,44 г. Ca +2H2O = Ca(OH)2 +H2 3. При действии цинка на соляную кислоту было получено 5 г. водорода. Найдите объем кислоты, вступившей в химическую реакцию. 4. Какой объем хлора прореагировал с иодидом калия, если при этом получилось 25,4 г. иода. 2KI + Cl2 = 2KCl + I2 5. Рассчитайте объем углекислого газа (н.у.), выделившегося при разложении 40 г. карбоната кальция.СаСО3 = СаО + СО2 6. Рассчитайте объем водорода образовавшегося при взаимодействии 6,5 г. цинка с соляной кислоты. Zn+2HCl=ZnCl2+H2 7. Найдите массу и количество вещества нитрата меди (II), образовавшегося при взаимодействии оксида меди (II) массой 4 г. с азотной кислотой. Сu +4 HNO3(конц. ) =Cu(NO3)2 + 2NO2 +2 H2O 8. Рассчитайте объем (н.у.) кислорода, который необходим для полного сгорания магния массой 4,8 г. 2Mg + O2 = 2MgO 9. Рассчитайте массу и объем углекислого газа , который образуется при разложении 50 г. карбоната кальция СаСО3 = СаО + СО2 10. Рассчитайте объем хлороводорода при нормальных условиях, полученный при действии серной кислоты на 5,85 г. хлорида натрия. H2SO4 + NaCl = HCl + NaHSO4 Хлороводород - HCl 11. Рассчитайте массу соли, которая образуется при взаимодействии 40 г. цинка с раствором хлороводорода. Zn+2HCl=ZnCl2+H2

Answer 1

Ответ:

1. 21.3gr

2.2,4gr

3, 11,2l

4.2,24l

5.

Answer 2

Mass of Sodium Sulfate formed:

To calculate the mass of sodium sulfate formed during the neutralization reaction between sodium hydroxide and sulfuric acid, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

Given: - Mass of sodium hydroxide (NaOH) = 12 g - Sulfuric acid (H2SO4) reacts with sodium hydroxide (NaOH) to form sodium sulfate (Na2SO4) and water (H2O)

The balanced chemical equation for the reaction is: 2NaOH + H2SO4 → Na2SO4 + 2H2O

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

1. Calculate the number of moles of sodium hydroxide (NaOH): - Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol - Number of moles of NaOH = Mass of NaOH / Molar mass of NaOH - Number of moles of NaOH = 12 g / 39.99 g/mol = 0.300 mol

2. Calculate the number of moles of sulfuric acid (H2SO4): - Molar mass of H2SO4 = 1.01 g/mol + 32.07 g/mol + (4 × 16.00 g/mol) = 98.09 g/mol - From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4. - Number of moles of H2SO4 = 0.300 mol / 2 = 0.150 mol

Since the stoichiometric ratio between NaOH and H2SO4 is 2:1, we can see that the number of moles of H2SO4 is half of the number of moles of NaOH. Therefore, H2SO4 is the limiting reactant.

Now, we can calculate the mass of sodium sulfate (Na2SO4) formed using the number of moles of H2SO4.

3. Calculate the number of moles of Na2SO4: - From the balanced equation, we can see that 1 mole of H2SO4 reacts to form 1 mole of Na2SO4. - Number of moles of Na2SO4 = Number of moles of H2SO4 = 0.150 mol

4. Calculate the mass of Na2SO4: - Molar mass of Na2SO4 = (2 × 22.99 g/mol) + 32.07 g/mol + (4 × 16.00 g/mol) = 142.04 g/mol - Mass of Na2SO4 = Number of moles of Na2SO4 × Molar mass of Na2SO4 - Mass of Na2SO4 = 0.150 mol × 142.04 g/mol = 21.31 g

Therefore, the mass of sodium sulfate formed during the neutralization reaction is 21.31 grams [[1]].

Mass of Calcium Hydroxide formed:

To calculate the mass of calcium hydroxide formed during the reaction between calcium and water, we need to determine the limiting reactant first.

Given: - Mass of water (H2O) = 4.44 g - Calcium (Ca) reacts with water (H2O) to form calcium hydroxide (Ca(OH)2) and hydrogen gas (H2)

The balanced chemical equation for the reaction is: Ca + 2H2O → Ca(OH)2 + H2

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

1. Calculate the number of moles of water (H2O): - Molar mass of H2O = 1.01 g/mol + (2 × 16.00 g/mol) = 18.02 g/mol - Number of moles of H2O = Mass of H2O / Molar mass of H2O - Number of moles of H2O = 4.44 g / 18.02 g/mol = 0.246 mol

2. Calculate the number of moles of calcium (Ca): - Molar mass of Ca = 40.08 g/mol - From the balanced equation, we can see that 1 mole of Ca reacts with 2 moles of H2O. - Number of moles of Ca = 0.246 mol / 2 = 0.123 mol

Since the stoichiometric ratio between Ca and H2O is 1:2, we can see that the number of moles of Ca is half of the number of moles of H2O. Therefore, Ca is the limiting reactant.

Now, we can calculate the mass of calcium hydroxide (Ca(OH)2) formed using the number of moles of Ca.

3. Calculate the number of moles of Ca(OH)2: - From the balanced equation, we can see that 1 mole of Ca reacts to form 1 mole of Ca(OH)2. - Number of moles of Ca(OH)2 = Number of moles of Ca = 0.123 mol

4. Calculate the mass of Ca(OH)2: - Molar mass of Ca(OH)2 = 40.08 g/mol + (2 × 16.00 g/mol) + (2 × 1.01 g/mol) = 74.10 g/mol - Mass of Ca(OH)2 = Number of moles of Ca(OH)2 × Molar mass of Ca(OH)2 - Mass of Ca(OH)2 = 0.123 mol × 74.10 g/mol = 9.12 g

Therefore, the mass of calcium hydroxide formed during the reaction is 9.12 grams [[2]].

Volume of Hydrochloric Acid reacted:

To find the volume of hydrochloric acid (HCl) that reacted with zinc (Zn) to produce 5 g of hydrogen gas (H2), we need to use the ideal gas law.

Given: - Mass of hydrogen gas (H2) = 5 g

The balanced chemical equation for the reaction is: Zn + 2HCl → ZnCl2 + H2

1. Calculate the number of moles of hydrogen gas (H2): - Molar mass of H2 = 2.02 g/mol - Number of moles of H2 = Mass of H2 / Molar mass of H2 - Number of moles of H2 = 5 g / 2.02 g/mol = 2.48 mol

2. Use the stoichiometry of the balanced equation to determine the number of moles of HCl: - From the balanced equation, we can see that 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of H2. - Number of moles of HCl = Number of moles of H2 / 1 = 2.48 mol

3. Calculate the volume of HCl using the ideal gas law: - The ideal gas law equation is: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. - At normal conditions (STP), the pressure (P) is 1 atm and the temperature (T) is 273.15 K. - The ideal gas constant (R) is 0.0821 L·atm/(mol·K). - Rearranging the ideal gas law equation to solve for volume (V), we have V = nRT / P. - Volume of HCl = Number of moles of HCl × R × T / P - Volume of HCl = 2.48 mol × 0.0821 L·atm/(mol·K) × 273.15 K / 1 atm - Volume of HCl = 56.7 L

Therefore, the volume of hydrochloric acid that reacted is 56.7 liters [[3]].

Volume of Chlorine reacted:

To find the volume of chlorine (Cl2) that reacted with potassium iodide (KI) to produce 25.4 g of iodine (I2), we need to use the stoichiometry of the balanced equation.

Given: - Mass of iodine (I2) = 25.4 g

The balanced chemical equation for the reaction is: 2KI + Cl2 → 2KCl + I2

1. Calculate the number of moles of iodine (I2): - Molar mass of I2 = 126.90 g/mol - Number of moles of I2 = Mass of I2 / Molar mass of I2 - Number of moles of I2 = 25.4 g / 126.90 g/mol = 0.200 mol

2. Use the stoichiometry of the balanced equation to determine the number of moles of Cl2: - From the balanced equation, we can see that 2 moles of KI react with 1 mole