Question

Рассчитайте объем , который будет занимать 64г оксида серы (4) и 64г кислорода (н.у)

Answer 1

M= 64(г)

n=m/M
M(SO2)=32+2*16=64 (г/моль)
n=64/64=1(моль)
V=Vm*n, где Vm - объем газа при н.у. =22.4(л/моль)
V=22.4*1=22.4(л)

m=64 (г)
M(O)=16(г/моль)
n=64/16=4(моль)
V=22.4*4=89.6(л)

Answer 2

Calculating the Volume of Sulfur Dioxide and Oxygen

To calculate the volume of 64g of sulfur dioxide (SO2) and 64g of oxygen (O2) at standard conditions, we can use the ideal gas law equation:

V = (nRT)/P

Where: - V = volume of the gas - n = number of moles of the gas - R = ideal gas constant - T = temperature in Kelvin - P = pressure

First, we need to find the number of moles of each gas using their respective molar masses.

The molar mass of sulfur dioxide (SO2) is 64g/mol, and the molar mass of oxygen (O2) is 32g/mol.

Calculating the Number of Moles

For sulfur dioxide: n(SO2) = mass / molar mass = 64g / 64g/mol = 1 mol

For oxygen: n(O2) = mass / molar mass = 64g / 32g/mol = 2 mol

Now, we can calculate the volume of each gas using the ideal gas law at standard conditions (0°C or 273.15K and 1 atm or 101.3 kPa).

Calculating the Volume of Sulfur Dioxide

Using the ideal gas law equation: V(SO2) = (nRT)/P - n = 1 mol - R = 0.0821 L·atm/(K·mol) (ideal gas constant) - T = 273.15K - P = 1 atm

Substituting the values: V(SO2) = (1 mol * 0.0821 L·atm/(K·mol) * 273.15K) / 1 atm = 22.4 L

Calculating the Volume of Oxygen

Using the ideal gas law equation: V(O2) = (nRT)/P - n = 2 mol - R = 0.0821 L·atm/(K·mol) (ideal gas constant) - T = 273.15K - P = 1 atm

Substituting the values: V(O2) = (2 mol * 0.0821 L·atm/(K·mol) * 273.15K) / 1 atm = 44.8 L

So, the volume of 64g of sulfur dioxide at standard conditions is 22.4 L, and the volume of 64g of oxygen at standard conditions is 44.8 L.