Question

Помогите решить Сколько литров O₂ вступит в реакцию с 13 граммами цинка Сколько грамм Na вступит в реакцию с 448 литрами O₂ Заранее благодарю !

Answer 1

       13 г                   Х л
1)    2Zn        +         O2        =        2ZnO

  n = 2 моль             n = 1 моль

Mr = 65 г/моль        Vm= 22,4 л/моль

m= 130 г                  V = 22,4 л

13 г Zn      -     Х л О2

130 г Zn    -    22,4 л О2

V(O2) = 13 * 22,4 / 130 = 2,24 л

         Х г                   448 л

2)      4Na          +      O2         =        2Na2O

      n = 4 моль           n = 1 моль

 Mr =23 г/моль          Vm = 22,4 л/моль

m =92 г                      V=22,4 л

Х г Na        -        448 л О2

92 г Na      -        22,4 л О2

m(Na) = 92 * 448 /22,4 =1840 г

 

Answer 2

Stoichiometry Calculations

To calculate the amount of O₂ that will react with 13 grams of zinc and the amount of Na that will react with 448 liters of O₂, we can use stoichiometry calculations based on the balanced chemical equations for the reactions.

Reaction 1: O₂ reacting with Zinc (Zn)

The balanced chemical equation for the reaction between O₂ and Zn is:

2 Zn + O₂ → 2 ZnO

From the balanced equation, we can see that 1 mole of O₂ reacts with 2 moles of Zn.

1. Calculate moles of Zn: - Moles of Zn = Mass of Zn / Molar mass of Zn - Moles of Zn = 13 g / 65.38 g/mol - Moles of Zn = 0.199 moles

2. Calculate moles of O₂: - Moles of O₂ = Moles of Zn / 2 (from the balanced equation) - Moles of O₂ = 0.199 moles / 2 - Moles of O₂ = 0.0995 moles

3. Convert moles of O₂ to liters at STP (Standard Temperature and Pressure): - 1 mole of any gas at STP occupies 22.4 liters - Liters of O₂ = Moles of O₂ * 22.4 - Liters of O₂ = 0.0995 moles * 22.4 - Liters of O₂ = 2.23 liters

Therefore, 2.23 liters of O₂ will react with 13 grams of zinc.

Reaction 2: O₂ reacting with Sodium (Na)

The balanced chemical equation for the reaction between O₂ and Na is:

4 Na + O₂ → 2 Na₂O

From the balanced equation, we can see that 1 mole of O₂ reacts with 4 moles of Na.

1. Convert liters of O₂ to moles at STP: - Moles of O₂ = Volume of O₂ / 22.4 (at STP) - Moles of O₂ = 448 L / 22.4 - Moles of O₂ = 20 moles

2. Calculate moles of Na: - Moles of Na = Moles of O₂ * 4 (from the balanced equation) - Moles of Na = 20 moles * 4 - Moles of Na = 80 moles

3. Calculate grams of Na: - Mass of Na = Moles of Na * Molar mass of Na - Mass of Na = 80 moles * 22.99 g/mol - Mass of Na = 1839.2 grams

Therefore, 1839.2 grams of Na will react with 448 liters of O₂.

I hope this helps! If you have any further questions, feel free to ask.