Question

Вычислите объем водорода, необходимого для реакции с кислородом объемом 5,6 л и (н.у.) .Какое количество вещества воды при этом образуется?

Answer 1

2H2 + O2=2H2O

n(O2)= 5,6 \ 22,4 = о,25 моль

Количество вещества водорода больше в 2 раза ,значит n=0,5 моль

Найдём объем воорода V(H2)= 0,5 * 22,4=11,2 литра

Answer 2

Calculation of Hydrogen Volume and Amount of Water Formed

To calculate the volume of hydrogen required for the reaction with 5.6 liters of oxygen at standard conditions (n.u.), we can use the ideal gas law equation:

PV = nRT

Where: - P is the pressure (at standard conditions, it is 1 atm) - V is the volume of the gas - n is the amount of substance in moles - R is the ideal gas constant (0.0821 L·atm/(mol·K)) - T is the temperature in Kelvin (at standard conditions, it is 273.15 K)

Since we are given the volume of oxygen (5.6 liters) and want to find the volume of hydrogen, we can assume that the reaction is balanced and that the volume ratio of hydrogen to oxygen is 2:1. This means that for every 2 liters of hydrogen, 1 liter of oxygen is required.

Let's calculate the volume of hydrogen required:

1. Convert the given volume of oxygen to liters: 5.6 liters 2. Divide the volume of oxygen by 2 to get the volume of hydrogen: 5.6 liters / 2 = 2.8 liters

Therefore, the volume of hydrogen required for the reaction is 2.8 liters.

To determine the amount of water formed in the reaction, we need to consider the balanced chemical equation for the reaction between hydrogen and oxygen, which 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.

To calculate the amount of water formed, we need to convert the volume of hydrogen to moles using the ideal gas law equation:

n = PV / RT

Let's calculate the amount of water formed:

1. Convert the volume of hydrogen to liters: 2.8 liters 2. Convert the volume of hydrogen to moles using the ideal gas law equation: - P = 1 atm - V = 2.8 liters - R = 0.0821 L·atm/(mol·K) - T = 273.15 K - n = (1 atm * 2.8 liters) / (0.0821 L·atm/(mol·K) * 273.15 K) = 0.112 moles of hydrogen

Since the reaction ratio is 2 moles of hydrogen to 2 moles of water, we can conclude that 0.112 moles of hydrogen will produce 0.112 moles of water.

Therefore, the amount of water formed in the reaction is 0.112 moles.

Please note that the calculations provided are based on the assumptions made and the given information.