question 2 of 15 if there are 5.40 moles of h, how many moles of each of the compounds are present? h₂so₄…

Explanation:

Step1: Determine mole - ratio for $H_2SO_4$

In $H_2SO_4$, there are 2 moles of H per 1 mole of $H_2SO_4$. Let $n_{H_2SO_4}$ be the moles of $H_2SO_4$ and $n_H$ be the moles of H. The mole - ratio is $\frac{n_{H_2SO_4}}{n_H}=\frac{1}{2}$. Given $n_H = 5.40$ mol, then $n_{H_2SO_4}=\frac{n_H}{2}$. $n_{H_2SO_4}=\frac{5.40}{2}=2.70$ mol

Step2: Determine mole - ratio for $C_2H_4O_2$

In $C_2H_4O_2$, there are 4 moles of H per 1 mole of $C_2H_4O_2$. Let $n_{C_2H_4O_2}$ be the moles of $C_2H_4O_2$. The mole - ratio is $\frac{n_{C_2H_4O_2}}{n_H}=\frac{1}{4}$. So $n_{C_2H_4O_2}=\frac{n_H}{4}$. $n_{C_2H_4O_2}=\frac{5.40}{4}=1.35$ mol

Step3: Determine mole - ratio for $NaOH$

In $NaOH$, there is 1 mole of H per 1 mole of $NaOH$. Let $n_{NaOH}$ be the moles of $NaOH$. The mole - ratio is $\frac{n_{NaOH}}{n_H}=\frac{1}{1}$. So $n_{NaOH}=n_H$. $n_{NaOH}=5.40$ mol

Answer:

$H_2SO_4$: 2.70 mol $C_2H_4O_2$: 1.35 mol $NaOH$: 5.40 mol