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 %
Answer:
q = -6464.9 kJ
Explanation:
We are given that the heat of combustion is ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.
vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³
m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g
mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol
q = −394 kJ /mol C x 16.41 mol C = -6464.9 kJ
Weight of sodium thiosulfate = 76.148 - 8.2
= 67.948 g.
Concentration of the solution = 67.948 / 172.7
= 0.393 g / mL. to the nearest thousandth . (answer).
Answer:
- chips n other food items (ones with salt)
- clorox
- aleve
- baking soda
3/5 times 5/3x = 8*3/5. X=24/5 simplified would be x= 4.8 L.
