Which of the following reactions is a neutralization reaction? A. ZnCl2(aq) + CaCrO4(aq) → ZnCrO4(s) + CaCl2(aq) B. HNO3(aq) + L
1 answer:
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The question is incomplete, here is the complete question:
How many grams of zinc would be required to produce 9.65g of zinc hydroxide
Zn+2MnO₂+H₂O→Zn(OH)₂+Mn₂O₃
<u>Answer:</u> The mass of zinc required is 6.35 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of zinc hydroxide = 9.65 g
Molar mass of zinc hydroxide = 99.4 g/mol
Putting values in equation 1, we get:
The given chemical equation follows:
By Stoichiometry of the reaction:
1 mole of zinc hydroxide is produced from 1 mole of zinc
So, 0.0971 moles of zinc hydroxide will be produced from = of zinc
Now, calculating the mass of zinc from equation 1, we get:
Molar mass of zinc = 65.4 g/mol
Moles of zinc = 0.0971 moles
Putting values in equation 1, we get:
Hence, the mass of zinc required is 6.35 grams
Answer:
Reagent O₂ will be consumed first.
Explanation:
The balanced reaction between O₂ and C₄H₁₀ is:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles
- O₂: 13 moles
- CO₂: 8 moles
- H₂O: 10 moles
Being:
- C: 12 g/mole
- H: 1 g/mole
- O: 16 g/mole
The molar mass of the compounds that participate in the reaction is:
- C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
- H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles* 58 g/mole= 116 g
- O₂: 13 moles* 32 g/mole= 416 g
- CO₂: 8 moles* 44 g/mole= 352 g
- H₂O: 10 moles* 18 g/mole= 180 g
If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?
mass of O₂= 223.78 grams
But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>