The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
<h3>What is Limiting reagent ?</h3>
The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.
Given chemical equation in balanced form ;
2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).
According to the Chemical equation ;
- The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.
But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
- 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react
As 80 g NaOH produces 106 g of Na₂CO₃.
Therefore 5 g NaoH will produce ;
106 / 80 x 5 = 6.625 g
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The ratios which are needed to determine the mass of oxygen produced from the decomposition of 10 grams of potassium chlorate are;
- 31.998 g O2 : 1 mole O2
- 3 mole O2 : 2 mole KClO3
- 112.55 g KClO31 mole KClO3
From stoichiometry;
- We can conclude that according to the reaction;
3 moles of oxygen requires 2 moles of KClO3 to be produced.
And from molar mass analysis;
- 31.998 g O2 is equivalent to 1 mole O2
- O2112.55 g KClO3 is equivalent to 1 mole KClO3
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According to the reaction equation:
by using ICE table:
CO(g) + Cl2(g) ↔ COCl2(g)
initial 0.025 0.0035 0
change -X -X +X
Equ (0.025-X) (0.0035-X) X
when:
Kp = P(COCl2)/P(CO}*P(Cl2) when Kp= 0.2
by substitution:
0.2 = X / (0.025-X)(0.0035-X) by solving this equation for X
∴ X = 1.74x10^-5
Combustion is one of the main ones, it is a chemical reaction in everyday life, because every time you strike a match or burn a candle that is the reaction of combustion.<span />
Hey there! :
Volume = 5.0 L
mass = 6.0 Kg
Therefore:
Density = mass / Volume
Density = 6.0 / 5.0
Density = 1.2 kg/L
