Answer:
The limiting reactant is oxygen gas and the reaction would produce 3.932 x 10 ^ 7 moles of water
Explanation:
Step 1: Convert everything into moles
nH2(l) = 1.06 x 10^8 g / 2.016 g/mol = 5.26 x 10^7 mols
nO2(l) = 6.29 x 10^8 g / 32.00 g/ mol = 1.966 x 10^7 mols
Step 2: Find the limiting reagent
The limiting reagent would be oxygen gas from
the balanced equation because we have less moles of oxygen gas needed to fully combust with the hydrant gas
Step 3: Stoichiometry time
The mole ratio from oxygen gas to water is 1:2
This means that for every mole of oxygen gas two moles of water is produced
We need to multiply the moles of oxygen gas by two to find out how many moles of water has been produced
nH2O = nO2 x 2
nH2O = 1.966x10^7 x 2
nH2O = 3.932x10^7
Step 4: Therefore statement
Therefore the limiting reactant is oxygen gas and the reaction would produce 3.932 x 10 ^ 7 moles of water
Answer:
1.788 C DEGREES
Explanation:
STP is 1 atm at 273.15 K
P1V1/T1 = P2V2/T2
(1)(62.65) / (273.15) = (612/760)(78.31)/T2
T2 = 274.93 K = 1.788 C
Answer:
6.05 g
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity
V = volume of solution in liter ,
n = moles of solute ,
From the question ,
M = 200mM
Since,
1 mM = 10⁻³ M
M = 200 * 10⁻³ M
V = 250 mL
Since,
1 mL = 10⁻³ L
V = 250 * 10⁻³ L
The moles can be calculated , by using the above relation,
M = n / V
Putting the respective values ,
200 * 10⁻³ M = n / 250 * 10⁻³ L
n = 0.05 mol
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From the question ,
m = 121 g/mol
n = 0.05 mol ( calculated above )
The mass of tri base can be calculated by using the above equation ,
n = w / m
Putting the respective values ,
0.05 mol = w / 121 g/mol
w = 0.05 mol * 121 g/mol
w = 6.05 g
Answer : The pH of buffer is 9.06.
Explanation : Given,
Concentration of HBrO = 0.34 M
Concentration of KBrO = 0.89 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KBrO]}{[HBrO]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKBrO%5D%7D%7B%5BHBrO%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of buffer is 9.06.
