What volume of 0.105 M HNO3, in milliliters, is required to react completely with 1.40 g of Ba(OH)2?
2 HNO3(aq) + Ba(OH)2(s) Ba(NO3)2(aq) + 2 H2O(l)
To solve this problem, we need to determine the amount of HNO3 required to react completely with 1.40 g of Ba(OH)2.
Step 1: Calculate the number of moles of Ba(OH)2.
To do this, we use the formula:
moles = mass / molar mass
The molar mass of Ba(OH)2 can be calculated by adding up the atomic masses:
Ba: 1 atomic mass unit
O: 16 atomic mass units (there are two oxygen atoms)
H: 1 atomic mass unit (there are two hydrogen atoms)
Molar mass of Ba(OH)2 = 137 + 16 + 1 + 1 + 16 + 1 = 171 g/mol
moles of Ba(OH)2 = 1.40 g / 171 g/mol ≈ 0.0082 mol
Step 2: Determine the mole ratio between HNO3 and Ba(OH)2.
From the balanced chemical equation, we can see that 2 moles of HNO3 react with 1 mole of Ba(OH)2.
Step 3: Calculate the moles of HNO3 required.
Using the mole ratio, we can calculate the moles of HNO3 required:
moles of HNO3 = 2 * moles of Ba(OH)2 = 2 * 0.0082 mol = 0.0164 mol
Step 4: Calculate the volume of the HNO3 solution.
Given that the concentration of HNO3 is 0.105 M, we can convert moles to volume using the formula:
moles = volume (in liters) * concentration
0.0164 mol = volume (in liters) * 0.105 M
Rearranging the equation: volume (in liters) = 0.0164 mol / 0.105 M
Since the volume is requested in milliliters, we need to convert liters to milliliters:
volume (in milliliters) = volume (in liters) * 1000
volume (in milliliters) = (0.0164 mol / 0.105 M) * 1000 ≈ 156.19 mL
Therefore, approximately 156.19 mL of 0.105 M HNO3 is required to react completely with 1.40 g of Ba(OH)2.
Convert 1.4 g Ba(OH)2 to moles. moles = grams/molar mass
Using the coefficients in the balanced equation, convert moles Ba(OH)2 to moles HNO3.
Finally, M = moles/L soln. You have M HNO3 and moles HNO3, solve for L.