Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
I need a little more context but I believe you are correct
There is no reaction.
<em>Molecular equation
:</em>
K₂CO₃(aq) + 2NH₄Cl(aq) ⟶ 2KCl(aq) + (NH₄)₂CO₃(aq)
<em>Ionic equation
:</em>
2K⁺(aq) + CO₃²⁻(aq) + 2NH₄⁺(aq) +2Cl⁻(aq) ⟶ 2K⁺(aq) + 2Cl⁻(aq) + 2NH₄⁺(aq) + CO₃²⁻(aq)
<em>Net ionic equation
:</em>
Cancel all ions that appear on both sides of the reaction arrow (underlined).
<u>2K⁺(aq)</u> + <u>CO₃²⁻(aq)</u> + <u>2NH₄⁺(aq</u>) +<u>2Cl⁻(aq)</u> ⟶ <u>2K⁺(aq)</u> + <u>2Cl⁻(aq</u>) + <u>2NH₄⁺(aq)</u> + <u>CO₃²⁻(aq)</u>
<em>All ions cancel</em>. There is no net ionic equation.
Answer:
This question appears incomplete
Explanation:
This question appears incomplete because of the absence of options. However, hydrogen is placed in group 1 because it has just one electron in it's outermost shell (which happens to be the only shell it has) just like every other group 1A/group 1 element. While helium is placed in group 8A/group 18 because it has a completely filled outermost shell (which is also the only shell it has) just like every other element in group 8A/group 18.