<span>3.68 liters
First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements:
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol
Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286
Looking at the balanced equation for the reaction which is
2 C4H10(g)+13 O2(g)→8 CO2(g)+10 H2O(l)
It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have:
0.037851286 mol * 4 = 0.151405143 mol
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) )
T = absolute temperature (23C + 273.15K = 296.15K)
So let's solve the formula for V and the calculate using known values:
PV = nRT
V = nRT/P
V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm)
V = (3.679338871 L*atm)/(1 atm)
V = 3.679338871 L
So the volume of CO2 produced will occupy 3.68 liters.</span>
It will become a solid or a gas
Weathering and erosion produce smaller rock particles which, when mixed with dust result in different types of soil.
Thank You!
PM me for more help!
Answer:
0.020 moles of 
can be formed
Explanation:
1. First determine the number of moles of LiOH.
Molarity is given by the following expression:
Solving for moles of solute:
moles of solute = M * Liters of solution
Converting 175.0mL to L:
Replacing values:
moles of solute = 0.227M*0.175L
moles of solute = 0.040
Therefore there are 0.040 moles of LiOH
2. Then write the balanced chemical reaction and use the stoichiometry of the reaction to calculate the number of moles of produced:
As the problem says that there are excess of 
, the limiting reagent is the LiOH.
can be formed
