consider the combustion reaction of acetylene (c₂h₂): 2c₂h₂ + 5o₂→4co₂ + 2h₂o. use the periodic table to…

Explanation:

Step1: Calculate molar mass of $C_2H_2$

The molar mass of $C$ is approximately $12.01\ g/mol$ and of $H$ is approximately $1.01\ g/mol$. For $C_2H_2$, $M_{C_2H_2}=2\times12.01 + 2\times1.01= 26.04\ g/mol$.

Step2: Calculate moles of $C_2H_2$

Given mass of $C_2H_2$ is $m = 859.0\ g$. Using the formula $n=\frac{m}{M}$, the number of moles of $C_2H_2$, $n_{C_2H_2}=\frac{859.0\ g}{26.04\ g/mol}\approx32.99\ mol$.

Step3: Determine mole - ratio from the balanced equation

From the balanced equation $2C_2H_2 + 5O_2\rightarrow4CO_2+2H_2O$, the mole - ratio of $C_2H_2$ to $O_2$ is $\frac{n_{C_2H_2}}{n_{O_2}}=\frac{2}{5}$.

Step4: Calculate moles of $O_2$

$n_{O_2}=\frac{5}{2}\times n_{C_2H_2}=\frac{5}{2}\times32.99\ mol = 82.475\ mol$.

Step5: Calculate mass of $O_2$

The molar mass of $O_2$ is $M_{O_2}=2\times16.00\ g/mol = 32.00\ g/mol$. Using $m = n\times M$, the mass of $O_2$, $m_{O_2}=82.475\ mol\times32.00\ g/mol = 2639.2\ g\approx2640.0\ g$.

Answer:

$2640.0\ g\ O_2$