How many moles of HNO3 are present if 0.236mol of Ba(OH)2 was needed to neutralize the acid solution?
2HNO3 + Ba(OH)2 ==> 2H2O x Ba(NO3)3
If we used 0.236 mols Ba(OH)2 = 2*0.236 mols HNO3.
Well, to find the number of moles of HNO3, we need to use the balanced equation for the reaction. But before we do that, let me take a moment to appreciate the fact that you just had a chemical reaction without even leaving your house. Talk about convenience!
Anyway, moving on to the calculation. The balanced equation shows that 2 moles of HNO3 react with 1 mole of Ba(OH)2. So, if 0.236 moles of Ba(OH)2 were needed to neutralize the acid solution, we can say that half of that number, which is 0.118 moles, is the number of moles of HNO3 present.
See? Chemistry can be as simple as splitting the difference!
To determine the number of moles of HNO3 present, we need to use the balanced chemical equation for the reaction between HNO3 and Ba(OH)2.
The balanced equation is as follows:
2HNO3 + Ba(OH)2 → Ba(NO3)2 + 2H2O
From the balanced equation, we can see that 2 moles of HNO3 react with 1 mole of Ba(OH)2. Therefore, the ratio of HNO3 to Ba(OH)2 is 2:1.
Given that 0.236 moles of Ba(OH)2 were needed to neutralize the acid solution, we can calculate the number of moles of HNO3 as follows:
Number of moles of HNO3 = (2/1) × (0.236 mol Ba(OH)2)
= 0.472 mol HNO3
Therefore, there are 0.472 moles of HNO3 present in the acid solution.
To determine the number of moles of HNO3 present, we need to use the stoichiometry of the balanced chemical equation for the reaction between HNO3 and Ba(OH)2.
The balanced chemical equation is:
2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O
From the balanced equation, we can see that 2 moles of HNO3 react with 1 mole of Ba(OH)2.
Given that 0.236 moles of Ba(OH)2 was needed to neutralize the acid solution, we can now use this information to determine the moles of HNO3.
Since the stoichiometry of the equation tells us that 2 moles of HNO3 react with 1 mole of Ba(OH)2, we can set up the following ratio:
2 moles of HNO3 / 1 mole of Ba(OH)2 = x moles of HNO3 / 0.236 moles of Ba(OH)2
By cross-multiplying and solving for x, we find:
x = (2 moles of HNO3 * 0.236 moles of Ba(OH)2) / 1 mole of Ba(OH)2
x = 0.472 moles of HNO3
Therefore, there are 0.472 moles of HNO3 present in the acid solution.