Natural gas is primarily composed of methane (CH4)
Natural gas is a naturally occurring hydrocarbon mixture which is primarily composed of Methane(CH4), but it also contains ethane,propane and heavier hydrocarbon. In addition it contain small amount of nitrogen, carbon dioxide,hydrogen sulfide and traces amount of water.
Answer:
1.) AgNO₃
2.) 0.563 moles AgBr
Explanation:
The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).
AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)
Molarity (M) = moles / liters
100 mL = 1 L
AgNO₃
45.0 mL / 100 = 45.0 L
1.25 M = ? moles / 0.450 L
? moles = 0.563 moles
KBr
75.0 mL / 100 = 0.750 L
0.800 M = ? moles / 0.750 L
? moles = 0.600 moles
In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.
Answer:
C₅ H₁₂ O
Explanation:
44 g of CO₂ contains 12 g of C
30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .
18 g of H₂O contains 2 g of hydrogen
14.8 g of H₂0 will contain 1.644 g of H .
total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O
gram of O = 2.22
moles of C, O, H in the given compound = 8.236 / 12 , 2.22 / 16 , 1.644 / 1
= .6863 , .13875 , 1.644
ratio of their moles = 4.946 : 1 : 11.84
rounding off to digits
ratio = 5 : 1 : 12
empirical formula = C₅ H₁₂ O
