use the following information to answer the next question. when reacted with potassium hydroxide, sulfuric…

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: $h_{2}so_{4(aq)}+koh_{(aq)}\rightarrow k_{2}so_{4(aq)}+h_{2}o_{(l)}$. 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

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: $h_{2}so_{4(aq)}+koh_{(aq)}\rightarrow k_{2}so_{4(aq)}+h_{2}o_{(l)}$. 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

Answer

Explanation:

Step1: Balance the chemical equation

$$H_2SO_4(aq)+2KOH(aq)\rightarrow K_2SO_4(aq) + 2H_2O(l)$$

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