Question

Какое количество брома возможно присоединить к 2,24 л ацетилена (н. у)? вычислите массу и число молей полученного продукта

Answer 1

C2H2 + 2Br2 = CHBr2 - CHBr2
n(C2H2) = 2,24/22,4=0,1 моль
по уравнению n(С2Н2) относится к n(Br2) в соотношении 1 : 2, значит 
n(Br2) = 0,1 * 2 = 0,2 моль
тоже по уравнению реакции n(С2Н2) относится к n(CHBr2 - CHBr2) находятся в соотношении 1 к 1, значит n(CHBr2 - CHBr2) тоже равно 0,1 моль
находим массу m(CHBr2 - CHBr2) = 0,1 × 346 = 34,6гр
М(C2H2Br4) = (12×2)+(1×2)+(80×4)=346

Answer 2

Calculation of the Amount of Bromine that can be Added to 2.24 L of Acetylene (STP)

To calculate the amount of bromine that can be added to 2.24 L of acetylene at standard temperature and pressure (STP), we need to consider the stoichiometry of the reaction between bromine and acetylene.

The balanced chemical equation for the reaction between bromine and acetylene is as follows:

C2H2 + Br2 -> C2H2Br2

From the balanced equation, we can see that one mole of acetylene reacts with one mole of bromine to produce one mole of 1,2-dibromoethane.

To calculate the amount of bromine that can be added, we need to convert the volume of acetylene to moles using the ideal gas law equation:

PV = nRT

Where: P = pressure (at STP, it is 1 atm) V = volume of gas (2.24 L) n = number of moles R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (at STP, it is 273.15 K)

Plugging in the values, we can calculate the number of moles of acetylene:

n = PV / RT n = (1 atm) * (2.24 L) / (0.0821 L·atm/(mol·K) * 273.15 K) n ≈ 0.105 moles of acetylene

Since the stoichiometry of the reaction is 1:1 between acetylene and bromine, the amount of bromine that can be added will also be 0.105 moles.

To calculate the mass of the product, 1,2-dibromoethane, we need to know the molar mass of the compound. The molar mass of C2H2Br2 is:

(2 * atomic mass of carbon) + (2 * atomic mass of hydrogen) + (2 * atomic mass of bromine) = (2 * 12.01 g/mol) + (2 * 1.01 g/mol) + (2 * 79.90 g/mol) ≈ 180.84 g/mol

Now we can calculate the mass of the product:

Mass = number of moles * molar mass Mass = 0.105 moles * 180.84 g/mol Mass ≈ 18.99 g

Therefore, the mass of the product, 1,2-dibromoethane, is approximately 18.99 grams.

To summarize: - The amount of bromine that can be added to 2.24 L of acetylene at STP is approximately 0.105 moles. - The mass of the product, 1,2-dibromoethane, is approximately 18.99 grams.

Please note that the calculations provided are based on the information given and may vary depending on the specific conditions and assumptions made.