To convert 3 mol H2O to grams, just multiply by the molar mass of H2O.
<span>3 mol H2O * 18 g H2O / 1 mol H2O = 54 g H2O
Hope this helped.</span>
Complete balanced equation: 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O
Ionized equation (with spectator ions):
2H⁺ + 2NO₃⁻ + Ca²⁺ + 2OH⁻ → Ca²⁺ + 2NO₃⁻ + 2H₂O
By eliminating the ions that are the same of both sides of the equation (spectator ions):
2H⁺ + 2OH⁻ → 2H₂O [Net Ionic Equation]
Answer:
392g sulfuric acid are produced
Explanation:
Based on the balanced equation:
2HCl + Na2SO4 → 2NaCl + H2SO4
<em>2 moles of HCl produce 1 mole of sulfuric acid</em>
<em />
To solve the problem we need to find the moles of sulfuric acid produced based on the chemical equation. Then, using its molar mass -<em>Molar mass H2SO4 = 98g/mol- </em>we can find the mass of sulfuric acid produced:
<em>Moles sulfuric acid:</em>
8mol HCl * (1mol H2SO4 / 2mol HCl) = 4 mol H2SO4
<em>Mass sulfuric acid:</em>
4mol H2SO4 * (98g / mol) =
392g sulfuric acid are produced
Answer:
1 .
2.
Explanation:
The more stable the ionic compound, the more is it lattice energy.
- The more the charge on the cation and the anion, the greater is the lattice energy.
- The less the size of the cation and the anion, the greater is the lattice energy.
Scandium oxide (
) is an oxide in which
behaves as cation and
behaves as anion.
The compounds which has higher lattice energy than scandium oxide are:
1 .
This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation
is smaller than
. Thus, this corresponds to higher lattice energy.
2.
This is because the charge on the cation
is greater than that of
and also the size of the cation
is smaller than
. Thus, this corresponds to higher lattice energy.
Answer:
Percentage yield = 85.2%
Explanation:
Given data:
Mass of Mg = 21.3 g
Actual yield of MgO = 30.2 g
Percentage yield = ?
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of Mg = mass/molar mass
Number of moles of Mg = 21.3 g / 24.3 g/mol
Number of moles of Mg = 0.88 mol
Now we will compare the moles of MgO with Mg.
Mg : MgO
2 : 2
0.88 : 0.88
Mass of MgO:
Mass of MgO= moles × molar mass
Mass of MgO= 0.88 mol × 40.3g/mol
Mass of MgO = 35.46 g
Actual yield of MgO = 30.2 g
Percentage yield:
Percentage yield = Actual yield/theoretical yield × 100
Percentage yield = 30.2 g/ 35.46 g × 100
Percentage yield = 85.2%