Solid iron is mixed with a solution of copper (I) nitrate to form iron (III) nitrate solution and metal copper. (what is the equ
1 answer:
You might be interested in
Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M
Answer:
17.65 grams of O2 are needed for a complete reaction.
Explanation:
You know the reaction:
4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O
First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:
- N: 14 g/mol
- H: 1 g/mol
- O: 16 g/mol
So, the molar mass of the compounds in the reaction is:
- NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
- O₂: 2*16 g/mol= 32 g/mol
- NO: 14 g/mol + 16 g/mol= 30 g/mol
- H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol
By stoichiometry, they react and occur in moles:
- NH₃: 4 moles
- O₂: 5 moles
- NO: 4 moles
- H₂O: 6 moles
Then in mass, by stoichiomatry they react and occur:
- NH₃: 4 moles*17 g/mol= 68 g
- O₂: 5 moles*32 g/mol= 160 g
- NO: 4 moles*30 g/mol= 120 g
- H₂O: 6 moles*18 g/mol= 108 g
Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?
mass of O₂≅17.65 g
<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>