Question

N(MgCo3)＝ 6 моль Найти V(Co2) и m(MgO)

Answer 1

Ответ:

MgCO3 ->(нагревание)-> MgO + CO2(выделяется)

Исходя из уравнения реакции n(MgCO3) = n (CO2) = 6 моль

V(CO2) = n*V=22.4*6=134.4 дм(куб)

Исходя из уравнения реакции n(MgCO3) = n(MgO) = 6 моль

M(MgO)=24+16= 40г/моль

m(MgO)=n*M= 40г/моль * 6 моль=240г

Answer 2

Ответ:

134,4 л

240 г

Объяснение:MgCO3=MgO+CO2

Вещества образуются в соотношении 1:1(количество каждого продукта равно количеству реагента)

V=V()*n=6*22,4=134,4 л

m=M(MgO)*n=40*6=240 г

Answer 3

To find the volume of CO2 (V(Co2)) and the mass of MgO (m(MgO)) produced from the given amount of MgCO3, we first need to balance the chemical equation for the decomposition of MgCO3.

The balanced chemical equation for the decomposition of MgCO3 is as follows:

MgCO3(s) → MgO(s) + CO2(g)

According to the equation, 1 mole of MgCO3 decomposes to produce 1 mole of MgO and 1 mole of CO2.

Given: N(MgCO3) = 6 moles

Since 1 mole of MgCO3 produces 1 mole of MgO, the moles of MgO produced will also be 6 moles (6 moles of MgCO3 decomposes to form 6 moles of MgO).

m(MgO) = moles of MgO * molar mass of MgO m(MgO) = 6 moles * molar mass of MgO

The molar mass of MgO is: Mg: 24.305 g/mol O: 15.999 g/mol

Molar mass of MgO = 24.305 g/mol + 15.999 g/mol = 40.304 g/mol

m(MgO) = 6 moles * 40.304 g/mol ≈ 241.824 g

Now, to find the volume of CO2 (V(Co2)), we need to apply the ideal gas law, which is given by:

PV = nRT

Where: P = Pressure of the gas (assumed constant at STP - standard temperature and pressure: 1 atm or 101.3 kPa) V = Volume of the gas (in liters) n = moles of the gas (in this case, n(CO2) = 6 moles) R = Ideal gas constant (0.0821 L.atm/mol.K) T = Temperature (in Kelvin, assumed to be 273 K at STP)

Rearranging the equation to solve for V(CO2):

V(CO2) = n(CO2) * R * T / P

V(CO2) = 6 moles * 0.0821 L.atm/mol.K * 273 K / 1 atm

V(CO2) ≈ 133.94 liters

So, the volume of CO2 produced (V(CO2)) is approximately 133.94 liters, and the mass of MgO (m(MgO)) formed is approximately 241.824 grams.