Answer:
11.58 g
Explanation:
First we convert the given masses of both reactants into moles, using their respective molar masses:
- 6.89 g CaO ÷ 56 g/mol = 0.123 mol CaO
- 13.79 g H₂O ÷ 18 g/mol = 0.766 mol H₂O
As<em> 1 H₂O mol reacts completely with 1 CaO mol,</em> and there are more H₂O moles than CaO moles, water is the non limiting reactant.
We calculate the mass of the reacting moles of water (the same number of CaO moles, 0.123):
- 0.123 mol H₂O * 18 g/mol = 2.21 g
Finally we <u>calculate how many grams of water remained after the reaction</u>:
- 13.79 g - 2.21 g = 11.58 g
The value of x : 2
<h3>Further explanation</h3>
Semi-normal Hydrochloric acid solution = 0.5 N = 0.5 M
for titration :
M₁V₁n₁=M₂V₂n₂(1=HCl,2=Na₂CO₃)
MW Na₂CO₃.xH₂O=143
MW Na₂CO₃.xH₂O = MW Na₂CO₃+ MW xH₂O = 143
MW Na₂CO₃ = 106 g/mol
MW xH₂O = 18x
Equation :
Answer:
3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate.
Explanation:
The balance chemical equation is :
Mass of barium hydroxide octahydrate = 6.5 g
Moles of barium hydroxide octahydrate =
According to reaction, 2 moles of ammonium thiocyanate reacts with1 mole of barium hydroxide octahydrate. The 0.020635 moles of barium hydroxide octahydrate will react with:
Mass of 0.04127 moles of ammonium thiocyanate;
3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate
The ammonium ion is NH4+ which should have a molecular weight of 18g/mol. Then, the amount of ammonium ion in mol should be <span>7.65g/ (18g/mol)= 0.425 mol.
The formula for ammonium sulfate would be </span>(NH4)2SO4. For every ammonium sulfate, you need 2 ammonium ion. Then the amount of ammonium sulfate in mol should be: 0.425 mol * (1/2)= 0.2125mol
The volume of the solution would be: 0.2125mol/ (0.812 mol/liter)= 0.26169l= 261.70 ml