If heat of combustion of acetylene C2H2 is -1301.1 kJ/mol then :

a) Write balanced thermochemical equation of acetylene combustion
b) If 0.25 mol C2H2 reacted according to the previous equation , and what is the emitted energy ?
c) how many grams needed to give off 3900 kJ energy ?

I think it's so :

a) 2C2H2 + 5O2 => 4CO2 + 2H2O + 1301.1kJ

b) 0.25mol * 1301.1 kJ = 325.27 kJ energy released

c) 1301.1 kJ/1mol = 3900 kJ/ x mol = 2.99mol
then X g = mol * molar mass of C2H2
X g = 2.99mol * 26 g/mol = 77.74 g

How much of this do you know how to do. Show your work and let me go through it.

All of this looks good to me except for the last part. Since the number of significant digits is 4 (at least I'm calling 3900 as 4 because I assume it is 3900.) then that 2.99 is really 2.997 ( or if not 4 s.f. for 3900 then it would be 2.997 and round to 3) which makes that 77.7 come out slightly higher as 77.9 or (78.0 if you use 3 x 26).

a) Sure! The balanced thermochemical equation for the combustion of acetylene (C2H2) is:

2 C2H2(g) + 5 O2(g) -> 4 CO2(g) + 2 H2O(g)

b) If 0.25 mol of C2H2 reacted, we can use the heat of combustion (-1301.1 kJ/mol) to calculate the emitted energy. Since the coefficient of C2H2 is 2 in the balanced equation, the energy emitted per mole of C2H2 is (-1301.1 kJ/mol) × 2 = -2602.2 kJ.

Therefore, for 0.25 mol C2H2, the emitted energy would be (-2602.2 kJ/mol) × 0.25 mol = -650.55 kJ. Note that the negative sign indicates that energy is being released.

c) To determine how many grams of acetylene are needed to give off 3900 kJ of energy, we can use the molar mass of C2H2.

The molar mass of C2H2 is approximately 26 g/mol (12 g/mol for carbon + 2 g/mol for each hydrogen).

To calculate the number of moles required to give off 3900 kJ, we divide it by the heat of combustion per mole:

Number of moles = 3900 kJ / (-1301.1 kJ/mol) ≈ -2.998 mol

Since we can't have a negative number of moles, we take the magnitude and multiply it by the molar mass:

Mass of C2H2 = | -2.998 mol | × 26 g/mol ≈ 77.95 g

So, approximately 77.95 grams of acetylene are needed to give off 3900 kJ of energy.

a) To write the balanced thermochemical equation for the combustion of acetylene (C2H2), we need to include both the reactants and the products. The combustion of acetylene involves reacting it with oxygen (O2) to form carbon dioxide (CO2) and water (H2O).

The balanced equation is as follows:
C2H2 + 5/2 O2 → 2 CO2 + H2O

b) To determine the emitted energy when 0.25 moles of C2H2 react, we can use the heat of combustion, which is given as -1301.1 kJ/mol.

Since 1 mole of C2H2 reacts to give off -1301.1 kJ of energy, we can calculate the energy released when 0.25 moles react as follows:

Energy released = (-1301.1 kJ/mol) x (0.25 mol) = -325.275 kJ

Therefore, the emitted energy is approximately -325.275 kJ.

c) To determine the mass of acetylene needed to release 3900 kJ of energy, we need to convert the given energy in kJ to moles and then use the molar mass of C2H2 to convert it to grams.

Step 1: Convert kJ to moles of C2H2
(-1301.1 kJ/mol) x (x mol) = -3900 kJ
x = (3900 kJ) / (-1301.1 kJ/mol) ≈ 2.99 mol

Step 2: Convert moles to grams
Using the molar mass of C2H2 (26 g/mol):
Mass of C2H2 = (2.99 mol) x (26 g/mol) ≈ 77.74 g

Therefore, approximately 77.74 grams of acetylene are needed to release 3900 kJ of energy.