The given question is incomplete. The complete question is :
Gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If 1.31g of water is produced from the reaction of 4.65g of butane and 10.8g of oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.
Answer: 28.0 %
Explanation:
To calculate the moles :
According to stoichiometry :
13 moles of 
require 2 moles of butane
Thus 0.34 moles of will require=
of butane
Thus is the limiting reagent as it limits the formation of product and butane is the excess reagent.
As 13 moles of give = 10 moles of 
Thus 0.34 moles of give =
of
Mass of 
The percent yield of water is 28.0 %
<span>We know the relation between heat and temperature follows the formula:
Q = Ce * m * (Tf-Ti) , where Q is heat transfer, Ce is specific heat, m is mass, Tf is final temperature and Ti is initial temperature.
Note that heat value is given in kj so we need to change to joules, then as heat is absorbed the final temperature will increase.
4,689 = 0.385 * 34.2 * (Tf-24)
Tf = 4,689 / (0.385 * 34.2) + 24 =380.12ÂşC</span>
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>
4.14x10^-3 per minute
First, calculate how many atoms of Cu-61 we initially started with by
multiplying the number of moles by Avogadro's number.
7.85x10^-5 * 6.0221409x10^23 = 4.7273806065x10^19
Now calculate how many atoms are left after 90.0 minutes by subtracting the
number of decays (as indicated by the positron emission) from the original
count.
4.7273806065x10^19 - 1.47x10^19 = 3.2573806065x10^19
Determine the percentage of Cu-61 left.
3.2573806065x10^19/4.7273806065x10^19 = 0.6890455577
The formula for decay is:
N = N0 e^(-λt)
where
N = amount left after time t
N0 = amount starting with at time 0
λ = decay constant
t = time
Solving for λ:
N = N0 e^(-λt)
N/N0 = e^(-λt)
ln(N/N0) = -λt
-ln(N/N0)/t = λ
Now substitute the known values and solve:
-ln(N/N0)/t = λ
-ln(0.6890455577)/90m = λ
0.372447889/90m = λ
0.372447889/90m = λ
0.00413830987 1/m = λ
Rounding to 3 significant figures gives 4.14x10^-3 per minute as the decay
constant.
3.125 i believe is the answer
