The gaseous hydrocarbon acetylene, C2H2, is used in welders' torches because of the large amount of heat released when acetylene burns with oxygen.

2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
How many grams of oxygen gas are needed for the complete combustion of 230. g of acetylene?
Answer in g.

5 moles O2= 2 moles ace

2.5 moles O2=230g/molassAce
moles O2= 230/2.5 * 1/molmassAcd
gramsO2=32*molesO2=32*230/2.5 * 1/molmassAce
work out the molemass Ace, it ought to be 28+2 grams/mole

To determine the amount of oxygen gas needed for the complete combustion of 230 g of acetylene, we can use the stoichiometric coefficients from the balanced chemical equation.

The balanced equation shows that for every 2 moles of C2H2, we need 5 moles of O2.

1 mole of C2H2 is equal to its molar mass (2 × 12.01 g/mol + 2 × 1.01 g/mol) = 26.04 g/mol.

First, we need to convert the mass of acetylene into moles:

230 g C2H2 × (1 mol C2H2 / 26.04 g C2H2) = 8.83 mol C2H2

Using the mole ratio from the balanced equation, we can determine the amount of oxygen gas needed:

8.83 mol C2H2 × (5 mol O2 / 2 mol C2H2) = 22.08 mol O2

Finally, we can convert the moles of O2 into grams:

22.08 mol O2 × (32.00 g O2 / 1 mol O2) = 706.56 g O2

Therefore, 706.56 grams of oxygen gas are needed for the complete combustion of 230 g of acetylene.

To determine the number of grams of oxygen gas needed for the complete combustion of 230 g of acetylene (C2H2), we can use the balanced chemical equation:

2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)

From the equation, we can see that the molar ratio between acetylene and oxygen gas is 2:5. This means that for every 2 moles of acetylene, we need 5 moles of oxygen gas.

To find the number of moles of acetylene, we divide the given mass by its molar mass. The molar mass of acetylene (C2H2) is calculated as follows:

(2 × atomic mass of carbon) + (2 × atomic mass of hydrogen)
(2 × 12.01 g/mol) + (2 × 1.01 g/mol) = 26.04 g/mol

Therefore, the number of moles of acetylene (C2H2) is:
230 g / 26.04 g/mol = 8.829 mol

Using the molar ratio from the balanced equation, we can calculate the number of moles of oxygen gas required:

(8.829 mol acetylene) × (5 mol O2 / 2 mol acetylene) = 22.073 mol O2

Finally, to find the mass of oxygen gas needed, we multiply the number of moles of oxygen gas by its molar mass. The molar mass of oxygen gas (O2) is 32.00 g/mol:

22.073 mol O2 × 32.00 g/mol = 706.351 g

Therefore, approximately 706.351 grams of oxygen gas are needed for the complete combustion of 230 grams of acetylene.