The combustion of ethyne, shown below unbalance, produces heat which can be used to weld metals: C2H2 (g) +502 (g) →2CO: (g) +H20 (g) AH reaction= -1544kJ
How much ethyne gas (in g) would you need to react with excess oxygen according to this reaction in order to raise the temperature of 325 g of high carbon steel from 165'C to its melting point, 1540 C? The heat capacity of high carbon steel is 0.490 J/g'C. (Assume a complete reaction and that all heat is transferred from the reaction to the metal with no loss.)
a. 7.37g
b. 1.84 g
c. 4.13 g
d 3.69 g

3.69 g

Explanation:

Given that:

The mass m = 325 g

The change in temperature ΔT = ( 1540 - 165)° C

= 1375 ° C

Heat capacity = 0.490 J/g°C

The amount of heat required:

q = mcΔT

q =  325 × 0.490 × 1375

q = 218968.75 J

q = 218.97 kJ

The equation for the reaction is expressed as:

Then,

1 mole of the ethyne is equal to 26 g of ethyne required for 1544 kJ heat.

Thus, for 218.97 kJ, the amount of ethyne gas required will be:

= 3.69 g

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.