Question

question 2 of 15
if there are 5.40 moles of h, how many moles of each of the compounds are present?
h₂so₄
mol
c₂h₄o₂
mol
naoh
mol

Explanation:

Step1: Determine H - atom ratio in $H_2SO_4$

Each $H_2SO_4$ has 2 H - atoms. Let $n_{H_2SO_4}$ be moles of $H_2SO_4$. Then $n_H = 2n_{H_2SO_4}$. Given $n_H=5.40$ mol, so $n_{H_2SO_4}=\frac{5.40}{2}=2.70$ mol.

Step2: Determine H - atom ratio in $C_2H_4O_2$

Each $C_2H_4O_2$ has 4 H - atoms. Let $n_{C_2H_4O_2}$ be moles of $C_2H_4O_2$. Then $n_H = 4n_{C_2H_4O_2}$. Given $n_H = 5.40$ mol, so $n_{C_2H_4O_2}=\frac{5.40}{4}=1.35$ mol.

Step3: Determine H - atom ratio in $NaOH$

Each $NaOH$ has 1 H - atom. Let $n_{NaOH}$ be moles of $NaOH$. Then $n_H=n_{NaOH}$. Given $n_H = 5.40$ mol, so $n_{NaOH}=5.40$ mol.

Answer:

1. For $H_2SO_4$: 2.70 mol
2. For $C_2H_4O_2$: 1.35 mol
3. For $NaOH$: 5.40 mol