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
The new volume of the bag will be 789.5 mL.
<u>Explanation:</u>
As per the Charles law, at constant pressure the volume of the gas is directly related to its temperature in Kelvin (K). That is as the temperature increases, the gas expands and vice-versa.
V1 = 250 ml
V2 = ?
T1 = 19° C
T2 = 60° C
Now we have to rewrite the equation to get the new volume as,
V2 = 
=
<em> </em>= 789.47 ≈ 789.5 ml.
So the new volume of the bag will be 789.5 mL.
Answer:
Volume occupied by 55.5 moles of water gas at STP = 1240 Liters.
Explanation:
1st keep in mind that 'volume' in the context of problems like this apply only to the gas phase form of the substance. Water in solid form (ice) or liquid form (liquid water) do not apply. Volume in this case is referred to as 'molar volume' and is a gas occupying 22.4 liters at STP conditions (0°C, 1Atm).
So, if 1 molar volume of water gas (steam) occupies 22.4 Liters at STP, then 55.5 moles of water gas occupies 22.4 Liters/mole x 55.5 moles = 1243.2 Liters, or 1240 Liters (3 sig. figs.). :-)
Okay so, 24.00 mL of a 0.25 M NaOH solution is titrated.
