Consider the combustion reaction for acetylene. 2C_(2)H_(2)(l)+5O_(2)(g)arrow 4CO_(2)(g)+2H_(2)O(g) If the acetylene tank contains 37.0 mol of C_(2)H_(2) and the oxygen tank contains 81.0 mol of O_(2) what is the limitin reactant for this reaction? O_(2) CO_(2) H_(2)O

To determine the limiting reactant, we need to compare the mole ratio of the reactants to the stoichiometric ratio in the balanced chemical equation.<br /><br />The balanced chemical equation for the combustion reaction of acetylene is:<br />$2C_{2}H_{2}(l)+5O_{2}(g)\rightarrow 4CO_{2}(g)+2H_{2}O(g)$<br /><br />From the equation, we can see that the stoichiometric ratio of $C_{2}H_{2}$ to $O_{2}$ is 2:5.<br /><br />Now, let's calculate the mole ratio of the reactants:<br />Mole ratio of $C_{2}H_{2}$ to $O_{2}$ = $\frac{37.0 \text{ mol}}{2} : \frac{81.0 \text{ mol}}{5} = 18.5 : 16.2$<br /><br />Since the mole ratio of $C_{2}H_{2}$ to $O_{2}$ is less than the stoichiometric ratio, $C_{2}H_{2}$ is the limiting reactant.<br /><br />Therefore, the limiting reactant for this reaction is $C_{2}H_{2}$.