consider the combustion reaction for acetylene.\n2c₂h₂(l) + 5o₂(g) → 4co₂(g) + 2h₂o(g)\nif the acetylene…

consider the combustion reaction for acetylene.\n2c₂h₂(l) + 5o₂(g) → 4co₂(g) + 2h₂o(g)\nif the acetylene tank contains 37.0 mol of c₂h₂ and the oxygen tank contains 81.0 mol of o₂, what is the limiting reactant for this reaction?\no c₂h₂\no o₂\no co₂\no h₂o

consider the combustion reaction for acetylene.\n2c₂h₂(l) + 5o₂(g) → 4co₂(g) + 2h₂o(g)\nif the acetylene tank contains 37.0 mol of c₂h₂ and the oxygen tank contains 81.0 mol of o₂, what is the limiting reactant for this reaction?\no c₂h₂\no o₂\no co₂\no h₂o

Answer

Explanation:

Step1: Determine mole - ratio from equation

The balanced equation is $2C_2H_2(l)+5O_2(g)\rightarrow4CO_2(g) + 2H_2O(g)$. The mole - ratio of $C_2H_2$ to $O_2$ is $\frac{n_{C_2H_2}}{n_{O_2}}=\frac{2}{5}$.

Step2: Calculate the amount of $O_2$ needed for given $C_2H_2$

Given $n_{C_2H_2}=37.0$ mol. The amount of $O_2$ required for complete reaction of $C_2H_2$ is $n_{O_2\ required}=\frac{5}{2}\times n_{C_2H_2}$. Substituting the value of $n_{C_2H_2}$, we get $n_{O_2\ required}=\frac{5}{2}\times37.0\ mol = 92.5$ mol.

Step3: Compare required and available $O_2$

The available amount of $O_2$ is $n_{O_2\ available}=81.0$ mol. Since $n_{O_2\ available}(81.0\ mol)<n_{O_2\ required}(92.5\ mol)$, oxygen is the limiting reactant.

Answer:

B. $O_2$