Answer
There is necessary a total of 30.1 g of oxygen gas $O_2$ to react with 9.80 g of $C_2H_2$.
Work Step by Step
1. Calculate the molar mass of $C_2H_2$:
$C: 12.01g * 2= 24.02g $
$H: 1.008g * 2= 2.016g $
24.02g + 2.016g = 26.04g
$ \frac{1 mole (C_2H_2)}{ 26.04g (C_2H_2)}$ and $ \frac{ 26.04g (C_2H_2)}{1 mole (C_2H_2)}$
2. The balanced reaction is:
$2 C_2H_2(g) + 5O_2(g) -^{\Delta}-\gt 4CO_2(g) +2 H_2O(g) + 2600 kJ$
According to the coefficients, the ratio of $C_2H_2$ to $O_2$ is 2 to 5:
$ \frac{ 5 moles(O_2)}{ 2 moles (C_2H_2)}$ and $ \frac{ 2 moles (C_2H_2)}{ 5 moles(O_2)}$
3. Calculate the molar mass for $O_2$::
$O: 16.00g * 2= 32.00g $
$ \frac{1 mole (O_2)}{ 32.00g (O_2)}$ and $ \frac{ 32.00g (O_2)}{1 mole (O_2)}$
4. Use the conversion factors to find the mass of $O_2$
$9.80g(C_2H_2) \times \frac{1 mole(C_2H_2)}{ 26.04g( C_2H_2)} \times \frac{ 5 moles(O_2)}{ 2 moles (C_2H_2)} \times \frac{ 32.00 g (O_2)}{ 1 mole (O_2)} = 30.1g (O_2)$
