Question

Вычислите обьем водорода (н.у.), полученного при взаимодействии 6,75 г. алюминия с избытком соляной кислоты.

Answer 1

2Al+6HCl=2AlCl3+3H2
n=m/M(Ar) M=Ar  Ar(Al)=27  M=27г/моль
n(Al)=6,75/27=0,25 моль
n(Al)/n(H2)=2/3, по условию реакции в 1,5 раза больше,
 след.,  n(H2)=0,25*1,5=0,375 моль
V=n*Vm
V(H2)=0,375*22,4=8,4л
Ответ:8,4л

Answer 2

Calculation of Hydrogen Volume

To calculate the volume of hydrogen gas produced when 6.75 grams of aluminum reacts with an excess of hydrochloric acid, we need to use the balanced chemical equation for the reaction. The balanced equation is as follows:

2Al + 6HCl -> 2AlCl3 + 3H2

From the balanced equation, we can see that 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 3 moles of hydrogen gas.

To find the number of moles of aluminum, we divide the given mass of aluminum by its molar mass. The molar mass of aluminum is 26.98 g/mol.

Number of moles of aluminum = mass of aluminum / molar mass of aluminum Number of moles of aluminum = 6.75 g / 26.98 g/mol

Now, we can use the mole ratio from the balanced equation to find the number of moles of hydrogen gas produced.

Number of moles of hydrogen gas = (Number of moles of aluminum) x (3 moles of hydrogen gas / 2 moles of aluminum)

Finally, we can use the ideal gas law to calculate the volume of hydrogen gas at standard temperature and pressure (STP). STP is defined as 0 degrees Celsius (273.15 Kelvin) and 1 atmosphere of pressure.

Volume of hydrogen gas = (Number of moles of hydrogen gas) x (22.4 L/mol) at STP

Let's calculate the volume of hydrogen gas using the given information.

Step 1: Calculate the number of moles of aluminum: Number of moles of aluminum = 6.75 g / 26.98 g/mol = 0.2501 mol

Step 2: Calculate the number of moles of hydrogen gas: Number of moles of hydrogen gas = 0.2501 mol x (3 mol H2 / 2 mol Al) = 0.3752 mol

Step 3: Calculate the volume of hydrogen gas at STP: Volume of hydrogen gas = 0.3752 mol x 22.4 L/mol = 8.4 L

Therefore, the volume of hydrogen gas obtained when 6.75 grams of aluminum reacts with an excess of hydrochloric acid is approximately 8.4 liters