Question

При пропускании водорода V=11,2 л через оксид меди II. Сколько граммов меди восстановилось?При горении 6 г магния образовался оксид магния. Какой объем кислорода затратился на горение?

Answer 1

Ответ:

Дано:

V(H2)=11,2 л

Найти:

m(Cu)

Решение:

CuO+H2=Cu+H2O

n(H2)=V/Vm=11,2 л/22,4 л/моль=0,5 моль

n(H2):n(Cu)=1:1

n(Cu)=0,5 моль

m(Cu)=n*M=0,5 моль*64 л/моль=32 г

Ответ: 32 г

Объяснение:

сорри надеюсь я правильно поняла и написала то)

Answer 2

Calculation of grams of copper reduced when hydrogen is passed through copper(II) oxide:

To calculate the grams of copper reduced when hydrogen is passed through copper(II) oxide, we need to use the balanced chemical equation for the reaction:

CuO + H2 → Cu + H2O

From the equation, we can see that 1 mole of copper(II) oxide (CuO) reacts with 1 mole of hydrogen (H2) to produce 1 mole of copper (Cu) and 1 mole of water (H2O).

To determine the grams of copper reduced, we need to know the molar mass of copper (Cu). The molar mass of copper is approximately 63.55 g/mol.

Given that the volume of hydrogen (V) is 11.2 L, we can use the ideal gas law to calculate the number of moles of hydrogen:

PV = nRT

Where: P = pressure (assume constant) V = volume of gas (11.2 L) n = number of moles of gas R = ideal gas constant T = temperature (assume constant)

Since the pressure and temperature are not given, we can assume they are constant. Therefore, we can rewrite the equation as:

n = PV / RT

Now we can calculate the number of moles of hydrogen.

Next, we need to determine the stoichiometric ratio between copper(II) oxide and copper. From the balanced chemical equation, we can see that 1 mole of copper(II) oxide produces 1 mole of copper.

Finally, we can calculate the grams of copper reduced by multiplying the number of moles of copper by the molar mass of copper.

Please provide the temperature and pressure values so that we can proceed with the calculation.