Calculate the mass of lithium hydroxide that must be added to 50ml of water so that it may react exactly with 5ml solution of 1.5M nitric acid.
2.416g
I don't get that. You should show your work so we can show you where you are making an error.
LiOH + HNO3 ==> LiNO3 + H2O
mols HNO3 = M x L = 1.5M x 0.005L = 0.0075 mols.
mols LiOH = 0.0075
g LiOH = mols x molar mass = 0.0075 x 23.95 = 0.1796 g LiOH which rounds to 0.18 g LiOH to two significant figures or to 0.2g to one s.f. I can't tell from your problem how many s.f. you have. The 5 mL limits the problem to 1 but you may have omitted the zero from 5.0 mL. The 1.5 limits the problem to 2.
To find the mass of lithium hydroxide (LiOH) needed to react with the nitric acid solution, we can follow these steps:
Step 1: Determine the balanced chemical equation for the reaction between lithium hydroxide and nitric acid.
Lithium hydroxide (LiOH) reacts with nitric acid (HNO3) to form lithium nitrate (LiNO3) and water (H2O):
2LiOH + 2HNO3 → LiNO3 + 2H2O
According to the balanced equation, one mole of lithium hydroxide reacts with two moles of nitric acid.
Step 2: Calculate the number of moles of nitric acid.
Given that the nitric acid solution is 1.5 M and the volume is 5 ml, we can calculate the number of moles using the equation:
moles of nitric acid = molarity × volume (in liters)
First, convert the volume of nitric acid from milliliters to liters:
Volume (in liters) = 5 ml ÷ 1000 = 0.005 L
Now calculate the number of moles of nitric acid:
moles of nitric acid = 1.5 M × 0.005 L = 0.0075 moles
Step 3: Determine the mole ratio between lithium hydroxide and nitric acid.
From the balanced equation, we can see that one mole of lithium hydroxide reacts with two moles of nitric acid. Therefore, the mole ratio is 1:2.
Step 4: Calculate the number of moles of lithium hydroxide needed.
Since the mole ratio of lithium hydroxide to nitric acid is 1:2, the number of moles of lithium hydroxide needed is twice the number of moles of nitric acid:
moles of lithium hydroxide = 2 × moles of nitric acid = 2 × 0.0075 moles = 0.015 moles
Step 5: Calculate the molar mass of lithium hydroxide.
The molar mass of lithium hydroxide can be calculated by adding up the atomic masses of each element:
Molar mass of LiOH = (molar mass of Li) + (molar mass of O) + (molar mass of H)
= (6.941 g/mol) + (16.00 g/mol) + (1.008 g/mol)
= 23.949 g/mol
Step 6: Calculate the mass of lithium hydroxide needed in grams.
mass of lithium hydroxide = moles of lithium hydroxide × molar mass of LiOH
= 0.015 moles × 23.949 g/mol
= 0.359 g
Therefore, approximately 0.359 grams of lithium hydroxide must be added to 50 ml of water to react with 5 ml of 1.5 M nitric acid solution.