Reactant C is the limiting reactant in this scenario.
Explanation:
The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.
Balanced chemical reaction is:
A + 2B + 3C → 2D + E
number of moles
A = 0.50 mole
B = 0.60 moles
C = 0.90 moles
Taking A as the reactant
1 mole of A reacted to form 2 moles of D
0.50 moles of A will produce
= 
thus 0.50 moles of A will produce 1 mole of D
Taking B as the reactant
2 moles of B reacted to form 2 moles of D
0.60 moles of B reacted to form x moles of D
= 
x = 2 moles of D is produced.
Taking C as the reactant:
3 moles of C reacted to form 2 moles of D
O.9 moles of C reacted to form x moles of D
= 
= 0.60 moles of D is formed.
Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.
The atom that is oxidized : Cr
The oxidizing agent : H₃PO₄
<h3>Further explanation</h3>
Reaction
2 H₃PO₄ (aq) + 2Cr(s) → 2 CrPO₄ (aq) + 3H₂(g)
Atoms undergoing a reduction reaction (decrease in oxidation number) and an oxidation reaction (increase in oxidation number)
H⁺(in H₃PO₄) =+1
H₂=0
Cr = 0
Cr³⁺(in CrPO₄ )
the oxidizing agent.⇒which undergoes a reduction reaction and oxidizes another compound/element : H₃PO₄
The reaction between NaOH and HCl is as follows
NaOH + HCl ---> NaCl + H₂O
for neutralisation, H⁺ ions react with an equivalent amount of OH⁻ ions.
Number of NaOH moles reacted = 0.270 M/1000 mL/L x 37 mL = 0.00999 mol
number of HCl moles reacted = 0.270 M/1000 mL x 27 mL = 0.00729 mol
HCl reacts with NaOH in 1:1 molar ratio
Number of excess NaOH moles remaining - 0.00999 - 0.00729 = 0.0027 mol
total volume of solution = 37 mL + 27 mL = 64 mL = 0.064 L
Since there's excess OH⁻ ions, we can calculate pOH value first
pOH = - log [OH⁻]
[OH⁻] = 0.0027 mol / 0.064 L = 0.042 mol/L
pOH = -log(0.042 M)
pOH = 1.37
by knowing pOH we can calculate pH using the following equation;
pH + pOH = 14
pH = 14 - 1.37
pH = 12.63
it has less tightly bound electrons, is able to lose electron easily as compare to metal B at it has 4 unpaired electron in 3d sub-shell.
Answer:
Balanced half-reaction: 
Explanation:
Balanced half-reaction for the reduction of gaseous oxygen to aqueous hydrogen peroxide in basic solution should be written in the following step-by-step manner:
Half-reaction: 
Balance H and O in basic medium: 
Balance charge: 
Balanced half-reaction: 