Question

Какой объем водорода необходимо для реакции с кислородом, если оброзовалась 7,2 грамм воды.

Answer 1

     Х л                                                                  7,2 г
    2H2               +              O2              ->            2H2O
n = 2 моль                                                        n = 2 моль
Vm=22,4 л/моль                                              М = 18 г/моль
V = 44,8 л                                                        m = 36 г

Х л Н2      -     7,2 г Н2О
44,8 л Н2  -    36 г Н2О
V(H2) = 44,8 * 7,2 / 36 = 8,96 л

Answer 2

Calculating the Volume of Hydrogen Gas Required for the Reaction

To calculate the volume of hydrogen gas required for the reaction with oxygen to form 7.2 grams of water, we can use the stoichiometry of the reaction between hydrogen and oxygen to water.

The balanced chemical equation for the reaction is: 2H2 + O2 → 2H2O

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

Calculation Steps

1. Calculate the moles of water formed: - Given mass of water = 7.2 grams - Using the molar mass of water (H2O), which is approximately 18.02 g/mol, we can calculate the moles of water formed.

Moles of water = Given mass of water / Molar mass of water Moles of water = 7.2 g / 18.02 g/mol Moles of water ≈ 0.3996 moles

2. Calculate the moles of hydrogen gas required: - From the balanced chemical equation, 2 moles of hydrogen gas react to form 2 moles of water.

Moles of hydrogen gas required = Moles of water formed / 2 Moles of hydrogen gas required ≈ 0.3996 moles / 2 Moles of hydrogen gas required ≈ 0.1998 moles

3. Convert moles of hydrogen gas to volume at standard temperature and pressure (STP): - At STP (0°C and 1 atm), 1 mole of any gas occupies approximately 22.4 liters.

Volume of hydrogen gas required = Moles of hydrogen gas required * 22.4 L/mol Volume of hydrogen gas required ≈ 0.1998 moles * 22.4 L/mol Volume of hydrogen gas required ≈ 4.4752 liters

Therefore, approximately 4.4752 liters of hydrogen gas are required for the reaction with oxygen to form 7.2 grams of water.

Cited Sources

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