He balanced equation for the reaction of acetylene, c2h2, and oxygen in an acetylene torch is 2 c2h2 + 5 o2 → 4 co2 + 2 h2o. in this reaction the number of grams of oxygen required to react with 0.13 g of acetylene is

0.4 g

Explanation:

1) Balanced chemical equation:

2) 2C2H2 + 5 O2 → 4CO2 + 2 H2O

3) mole ratios:

2 mol C2h2 : 5 mol O2

4) Convert 0.13 g C2H2 into number of moles

n = mass in grams / molar mass

molar mass C2H2 = 2 *12g/mol + 2*1 g/mol = 26 g/mol

n = 0.13 g / 26 g/mol = 0.005 mol

5) Set the proportion with the unknown

5 mol O2             x
=
2 mol C2H2      0.005 mol C2H2

x = 0.005 mol C2H2 * 5 mol O2 / 2 mol C2H2 = 0.0125 mol O2

6) Convert 0.0125 mol O2 to grams

mass = number of moles * molar mass

molar mass of O2 = 32 g/mol

mass = 0.0125 mol * 32 g/mol = 0.4 g

0.4 g

30litre of li2o is 1.0002mole approximately 1 mole