1. how many grams of KOH will neutralize (react with) 0.15 mol HNO3
2. how many liters of 1 M HCl will neutralize 10 ML of 9.2 M NaOH
KOH + HNO3 ==> KNO3 + H2O
so you know it's 1 mol KOH to 1 mol HNO3.
mols HNO3 = M x L = ?
mols KOH = mols HNO3
mols KOH = grams/molar mass. You know mols and molar mass, solve for grams.
HCl + NaOH ==> NaCl + H2O
again 1:1
mols NaOH = M x L = ?
mols HCl = mols NaOH
M HCl = mols/L. You know M and mols, solve for L.
To find the answers to these questions, we need to use the concept of stoichiometry and the balanced chemical equations relating the reactants and products. Let's break down each question and explain how to solve it step by step:
1. How many grams of KOH will neutralize (react with) 0.15 mol HNO3?
To answer this question, we need to use the balanced chemical equation for the neutralization reaction between KOH and HNO3:
2 KOH + HNO3 -> KNO3 + H2O
From the equation, we can see that 2 moles of KOH react with 1 mole of HNO3. Therefore, we can write the following ratio:
2 mol KOH / 1 mol HNO3
Now, we can set up a proportion using the given 0.15 mol HNO3:
2 mol KOH / 1 mol HNO3 = x g KOH / 0.15 mol HNO3
Solving for x gives us:
x = (2 mol KOH / 1 mol HNO3) * 0.15 mol HNO3
x = 0.3 mol KOH
Finally, we can convert the moles of KOH to grams using its molar mass (39.10 g/mol):
Mass of KOH = 0.3 mol KOH * 39.10 g/mol = 11.73 g
Therefore, you would need approximately 11.73 grams of KOH to neutralize 0.15 mol of HNO3.
2. How many liters of 1 M HCl will neutralize 10 mL of 9.2 M NaOH?
To solve this question, we can again use the stoichiometry and balanced chemical equation, this time for the neutralization reaction between HCl and NaOH:
HCl + NaOH -> NaCl + H2O
From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH.
To find the number of moles of NaOH in 10 mL of a 9.2 M solution, we can use the following formula:
Moles = Volume (in L) * Concentration (in M)
Moles of NaOH = 10 mL * (9.2 mol/L / 1000 mL/L)
Moles of NaOH = 0.092 mol
Since the reaction is 1:1, we can conclude that 0.092 mol of HCl will be required to neutralize 0.092 mol of NaOH.
Now, since we know the concentration of HCl is 1 M, we can use the formula:
Volume (in L) = Moles / Concentration (in M)
Volume of HCl = 0.092 mol / 1 mol/L
Volume of HCl = 0.092 L or 92 mL
Therefore, approximately 0.092 liters (or 92 mL) of 1 M HCl will be required to neutralize 10 mL of 9.2 M NaOH.