I believe the answer is A the 1st one
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.
Answer:
The answer is 3
C2H5OH + O2 CO2 +H2O (unbalanced)
C2H5OH +3O2(g). 2CO2(g)+3H2O(balanced)
There are 19.5 g Na in 71.4 g NaHCO₃
Calculate the <em>molecular mass of NaHCO₃</em>.
1 Na = 1 × 22.99 u = 22.99 u
1 H = 1 × 1.008 u = 1.008 u
1 C = 1 × 12.01 u = 12.01 u
3 O = 3 × 16.00 u = <u>48.00 u
</u>
TOTAL = 84.008 u
So, there are 22.99 g of Na in 84.008 g NaHCO₃.
∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na