For the neutralization reaction involving HNO3 and Ca(OH)2, how many liters of 1.55 M HNO3 are needed to react with 45.8 mL of a 4.66 M Ca(OH)2 solution?
1. 0.137 L
2. 0.0343 L
3. 0.275 L
4. 1.32 L
5. 0.662 L
6. 0.330 L
If you're just trying to finish UT like this.
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To determine the amount of HNO3 needed, we can use the concept of stoichiometry and the formula:
M1V1 = M2V2
Where:
M1 = initial molarity of the HNO3 solution
V1 = initial volume of the HNO3 solution
M2 = final molarity of the Ca(OH)2 solution
V2 = final volume of the Ca(OH)2 solution
Given:
M1 = 1.55 M (molarity of HNO3)
V1 = ? (unknown volume of HNO3, which we need to find)
M2 = 4.66 M (molarity of Ca(OH)2)
V2 = 45.8 mL = 0.0458 L (volume of Ca(OH)2)
Rearranging the formula, we have:
V1 = (M2 × V2) / M1
Substituting the given values into the equation:
V1 = (4.66 M × 0.0458 L) / 1.55 M
Calculating this expression:
V1 = 0.137 L
Therefore, the number of liters of 1.55 M HNO3 needed to react with 45.8 mL of a 4.66 M Ca(OH)2 solution is 0.137 L.
So, the correct answer is option 1: 0.137 L.
2HNO3 + Ca(OH)2 ==> Ca(NO3)2 + 2H2O
mols Ca(OH)2 = M x L = ?
Using the coefficients in the balanced equation, convert mols Ca(OH)2 to mols HNO3.
Then M HNO3 = mols HNO3/LHNO3. You have mols and M, solve for L.