The mass of hydrated salt - 2.123 g
mass of anhydrous salt - 1.861 g
mass that has been reduced is the mass of water that has been heated and lost from the compound thereby making the salt anhydrous.
therefore mass of water lost - 2.123 - 1.861 = 0.262 g
number of moles of water lost - 0.262 g / 18 g/mol = 0.0146 mol
number of moles of salt - 1.861 g / 380.6 g/mol = 0.00490 mol
molar ratio of moles of water to moles of salt
molar ratio = 0.146 mol / 0.00490 mol = 2.98 rounded off to 3
for every 1 mol of salt there are 3 moles of water
therefore empirical formula - Cu₃(PO₄)₂.3H₂O
Answer:
209.98 g of NaOH
Explanation:
We are given;
- Volume of HCl as 3 L
- Molarity of HCl as 1.75 M
We are required to calculate the mass of NaOH required to completely neutralize the acid given.
First, we write a balanced equation for the reaction between NaOH and HCl
That is;
NaOH + HCl → NaCl + H₂O
Second, we determine the number of moles of HCl
Number of moles = Molarity × Volume
= 1.75 M × 3 L
= 5.25 moles
Third, we use the mole ratio to determine the moles of NaOH
From the reaction,
1 mole of NaOH reacts with 1 mole of HCl
Therefore;
Moles of NaOH = Moles of HCl
= 5.25 moles
Fourth, we determine the mass of NaOH
Molar mass of NaOH = 39.997 g/mol
Mass of NaOH = 5.25 moles × 39.997 g/mol
= 209.98 g
Thus, 209.98 g of NaOH will completely neutralize 3L of 1.74 M HCl
