Answer:
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Answer:
The answer to your question is below
Explanation:
1)
Balanced chemical reaction
2CH₃OH + 3O₂ ⇒ 2 CO₂ + 4H₂O
Reactant Element Product
2 C 2
8 H 8
8 O 8
Molar mass of CH₃OH = 2[12 + 16 + 4]
= 2[32]
= 64 g
Molar mass of O₂ = 3[16 x 2] = 96 g
Theoretical proportion CH₃OH/O₂ = 64 g/96g = 0.67
Experimental proportion CH₃OH/O₂ = 60/48 = 1.25
Conclusion
The limiting reactant is O₂ because the Experimental proportion was higher than the theoretical proportion
2)
Balanced chemical reaction
S₈ + 12O₂ ⇒ 8SO₃
Reactant Elements Products
8 S 8
24 O 24
Molar mass of S₈ = 32 x 8 = 256 g
Molar mass of O₂ = 12 x 32 = 384 g
Theoretical proportion S₈ / O₂ = 256 / 384
= 0.67
Experimental proportion S₈ / O₂ = 40 / 35
= 1.14
Conclusion
The limiting reactant is O₂ because the experimental proportion was lower than the theoretical proportion.
Answer:
450g of coke (C)
Explanation:
Step 1:
The balanced equation for the reaction is given below:
3C(s) + 2SO2(g) —> CS2(s) + 2CO2(g)
Step 2:
Determination of the mass of C that reacted and the mass of CS2 produced from the balanced equation.
This is illustrated below:
Molar Mass of C = 12g/mol
Mass of C from the balanced equation = 3 x 12 = 36g
Molar Mass of CS2 = 12 + (32x2) = 12 + 64 = 76g/mol.
From the balanced equation above, 36g of C reacted to produce 76g of CS2.
Step 3:
Determination of the mass of C required to produce 950g of CS2. This is illustrated below:
From the balanced equation above, 36g of C reacted to produce 76g of CS2.
Therefore, Xg of C will react to produce 950g of CS2 i.e
Xg of C = (36 x 950)/76
Xg of C = 450g
From the calculations made above, 450g of coke (C) is needed to produce 950g of CS2.
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