salicylic acid ($c_7h_6o_3$) reacts with acetic anhydride ($c_4h_6o_3$) to form acetylsalicylic acid…

salicylic acid ($c_7h_6o_3$) reacts with acetic anhydride ($c_4h_6o_3$) to form acetylsalicylic acid ($c_9h_8o_4$).\n\n$2c_7h_6o_3(aq)+c_4h_6o_3(aq)\rightarrow2c_9h_8o_4(aq)+h_2o(l)$\n\nwhat is the limiting reactant if 70.0 g of $c_7h_6o_3$ and 80.0 g of $c_4h_6o_3$ react?\n\no water\no salicylic acid\no acetic anhydride\no acetylsalicylic acid

Answer

Explanation:

Step1: Calculate molar masses

The molar mass of $C_7H_6O_3$: $M_{C_7H_6O_3}=(7\times12.01 + 6\times1.01+3\times16.00)\text{ g/mol}=138.12\text{ g/mol}$ The molar mass of $C_4H_6O_3$: $M_{C_4H_6O_3}=(4\times12.01 + 6\times1.01+3\times16.00)\text{ g/mol}=102.09\text{ g/mol}$

Step2: Calculate moles of reactants

The moles of $C_7H_6O_3$, $n_{C_7H_6O_3}=\frac{m}{M}=\frac{70.0\text{ g}}{138.12\text{ g/mol}}\approx0.507\text{ mol}$ The moles of $C_4H_6O_3$, $n_{C_4H_6O_3}=\frac{m}{M}=\frac{80.0\text{ g}}{102.09\text{ g/mol}}\approx0.784\text{ mol}$

Step3: Use stoichiometry

From the balanced - equation $2C_7H_6O_3(aq)+C_4H_6O_3(aq)\rightarrow2C_9H_8O_4(aq)+H_2O(l)$, the mole ratio of $C_7H_6O_3$ to $C_4H_6O_3$ is $2:1$. For $0.784\text{ mol}$ of $C_4H_6O_3$, the moles of $C_7H_6O_3$ required is $n = 2\times0.784\text{ mol}=1.568\text{ mol}$ But we have only $0.507\text{ mol}$ of $C_7H_6O_3$. So, $C_7H_6O_3$ will be consumed first.

Answer:

salicylic acid