Question

На 10 г суміші Cu i Al подіяли розчином H2SO4 при цьому виділилося 5.6 л газу. Визначити масові частки металів в суміші. ПОМОГИТЕ!!! ​

Answer 1

Ответ:

w(Al)=45%

w(Cu)=55%

Объяснение:

В реакцию вступает только алюминий:

2Al+3H2SO4=Al2(SO4)3+3H2

Количество водорода

n(H2)=V/Vm=5,6/22,4=0,25 моль

Количество алюминия по уравнению

n(Al)=0,25*2/3=0,1667 моль

Масса алюминия

m(Al)=n*M=0,1667*27=4,5 г

Массовая доля алюминия

w(Al)=4,5*100/10=45%

Массовая доля меди

w(Cu)=100-45=55%

Answer 2

Calculation of Mass Fractions of Cu and Al in the Mixture

To determine the mass fractions of Cu and Al in the mixture, we can use the concept of stoichiometry and the ideal gas law. The balanced chemical equation for the reaction between Cu and Al with H2SO4 is:

2Al + 3Cu + 8H2SO4 -> Al2(SO4)3 + 3CuSO4 + 8H2O + 5.6L of gas

From the balanced equation, we can see that the stoichiometric ratio between Cu and Al is 3:2. This means that for every 3 moles of Cu, we have 2 moles of Al.

Let's assume the mass of the mixture is M grams, and the mass of Cu and Al in the mixture are mCu and mAl, respectively. The mass fractions of Cu and Al can be calculated using the following formulas:

Mass fraction of Cu (wCu) = mCu / M Mass fraction of Al (wAl) = mAl / M

To find the mass fractions, we need to determine the moles of Cu and Al in the reaction mixture.

Calculation of Moles of Cu and Al

To calculate the moles of Cu and Al, we can use their molar masses and the given volume of gas produced.

The molar mass of Cu is 63.55 g/mol, and the molar mass of Al is 26.98 g/mol.

The volume of gas produced is given as 5.6 L.

Using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature, we can calculate the moles of gas produced.

Assuming the reaction takes place at standard temperature and pressure (STP), which is 0 degrees Celsius and 1 atmosphere, we can use the following values:

P = 1 atm V = 5.6 L R = 0.0821 L·atm/(mol·K) T = 273 K

Using the ideal gas law equation, we can calculate the moles of gas produced:

n = PV / RT

Substituting the values, we get:

n = (1 atm * 5.6 L) / (0.0821 L·atm/(mol·K) * 273 K)

Calculating this expression gives us the moles of gas produced.

Calculation of Mass of Cu and Al

Now that we have the moles of gas produced, we can calculate the moles of Cu and Al using the stoichiometric ratio.

For every 5.6 L of gas produced, we have 3 moles of Cu and 2 moles of Al.

Using these ratios, we can calculate the moles of Cu and Al:

moles of Cu = (moles of gas produced) * (3 moles of Cu / 5.6 L) moles of Al = (moles of gas produced) * (2 moles of Al / 5.6 L)

Calculation of Mass Fractions of Cu and Al

Now that we have the moles of Cu and Al, we can calculate their masses using their molar masses.

mass of Cu = (moles of Cu) * (molar mass of Cu) mass of Al = (moles of Al) * (molar mass of Al)

Finally, we can calculate the mass fractions of Cu and Al using the formulas mentioned earlier:

Mass fraction of Cu (wCu) = mass of Cu / mass of mixture Mass fraction of Al (wAl) = mass of Al / mass of mixture

By substituting the calculated values, we can determine the mass fractions of Cu and Al in the mixture.

Please note that the actual calculations require the values of the molar masses of Cu and Al, as well as the temperature and pressure conditions under which the reaction occurred. These values were not provided in the question, so the calculations cannot be performed accurately without this information.