Question

use the following information to answer the next question. when reacted with potassium hydroxide, sulfuric acid undergoes a neutralization reaction according to the following unbalanced equation: $h_{2}so_{4(aq)}+koh_{(aq)}
ightarrow k_{2}so_{4(aq)}+h_{2}o_{(l)}$. 6. the number of moles of potassium sulfate produced when 304 g of potassium hydroxide reacts with excess sulfuric acid is: 5.42, 8.53×10³, 10.8, 2.71

Explanation:

Step1: Balance the chemical equation

$$H_2SO_4(aq)+2KOH(aq) ightarrow K_2SO_4(aq) + 2H_2O(l)$$

Step2: Calculate the molar - mass of KOH

The molar - mass of K (potassium) is approximately $39.1\ g/mol$, O (oxygen) is approximately $16\ g/mol$, and H (hydrogen) is approximately $1.01\ g/mol$. So, the molar - mass of $KOH$ is $39.1+16 + 1.01=56.11\ g/mol$.

Step3: Calculate the number of moles of KOH

The number of moles of a substance is given by the formula $n=\frac{m}{M}$, where $m$ is the mass and $M$ is the molar - mass. Given $m = 304\ g$ of $KOH$, then $n_{KOH}=\frac{304\ g}{56.11\ g/mol}\approx5.42\ mol$.

Step4: Determine the mole ratio between KOH and $K_2SO_4$

From the balanced chemical equation, the mole ratio of $KOH$ to $K_2SO_4$ is $2:1$.

Step5: Calculate the number of moles of $K_2SO_4$

Let $n_{K_2SO_4}$ be the number of moles of $K_2SO_4$. Using the mole ratio, $n_{K_2SO_4}=\frac{1}{2}n_{KOH}$. Substituting $n_{KOH}\approx5.42\ mol$, we get $n_{K_2SO_4}=\frac{5.42\ mol}{2}=2.71\ mol$.

Answer:

2.71