Question

Какой обьём водорода (л) при н.у.потребуется для восстановления 800 г оксида меди (II), содержащего 40 % примесей?

Answer 1

Решение
10       х 
моль моль
CuO + H2 = Cu + H2O
  1         1
моль   моль
М(CuO)=64+16=80 г/моль
n(CuO)=m/M=800/80=10 моль
n(H2)=10*1/1=10 моль
V(H2)=Vm*n=22.4*10=224 дм3
Найдем объем водорода без примесей
224 -100%
 х    - 40%
х=224*40/100=89,6 дм3 составляют примеси
V(H2)=224-89,6=134,4 дм3
Ответ:V(H2)=134,4 дм3

Answer 2

Дано:
m(CuO) = 800г
w(CuO) = 0,6
Найти:
V(H2) = ?

Решение:
CuO + H2 = Cu + H2O
m(CuO) = 800*0,6 =480г
n(CuO) = m/M = 480/80 = 6 моль
n(H2) = 6 моль
V(H2) = Vm*n = 22,4*6 = 134,4л

Answer 3

Calculating the Volume of Hydrogen Gas Required to Reduce Copper(II) Oxide

To calculate the volume of hydrogen gas required to reduce 800 g of copper(II) oxide (CuO) containing 40% impurities, we can use the stoichiometry of the reaction between copper(II) oxide and hydrogen gas.

The balanced chemical equation for the reaction is: CuO + H2 → Cu + H2O

From the balanced equation, we can see that 1 mole of copper(II) oxide reacts with 1 mole of hydrogen gas to produce 1 mole of copper and 1 mole of water.

Step 1: Calculate the Moles of Copper(II) Oxide

First, we need to calculate the moles of copper(II) oxide in 800 g of the compound. We can use the molar mass of CuO to convert the mass to moles.

The molar mass of CuO is approximately 79.55 g/mol (copper: 63.55 g/mol, oxygen: 16.00 g/mol).

Moles of CuO = Mass of CuO / Molar mass of CuO Moles of CuO = 800 g / 79.55 g/mol

Moles of CuO = 10.06 moles

Step 2: Calculate the Moles of Hydrogen Gas

Since the stoichiometry of the reaction is 1:1 between CuO and H2, the moles of hydrogen gas required will be the same as the moles of copper(II) oxide.

Moles of H2 = Moles of CuO Moles of H2 = 10.06 moles

Step 3: Convert Moles of Hydrogen Gas to Volume at Standard Temperature and Pressure (STP)

At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.

Volume of H2 at STP = Moles of H2 * 22.4 L/mol Volume of H2 at STP = 10.06 moles * 22.4 L/mol

Volume of H2 at STP = 225.184 L

Therefore, approximately 225.184 liters of hydrogen gas at standard temperature and pressure (STP) are required to completely reduce 800 g of copper(II) oxide containing 40% impurities.

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