<span>Answer:
2 C8H18 + 25 O2 => 16 CO2 + 18 H2O
1.0 kg = 1000 g C8H18 = 1000 g / 114.2293 g/mole = 8.75 moles C8H18
8.75 moles C8H18 produce (16/2) (8.75) = 70 moles CO2
70 moles CO2 = (70 moles) (44.0096 g/mole) = 3081 g CO2 = 3.1 kg CO2</span>
Silver chloride produced : = 46.149 g
Limiting reagent : CuCl2
Excess remains := 3.74 g
<h3>Further explanation</h3>
Reaction
silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate
Required
silver chloride produced
limiting reagent
excess remains
Solution
Balanced equation
2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)
mol AgNO3 :
= 58.5 : 169,87 g/mol
= 0.344
mol CuCl2 :
=21.7 : 134,45 g/mol
= 0.161
mol ratio : coefficient of AgNO3 : CuCl2 :
= 0.344/2 : 0.161/1
= 0.172 : 0.161
CuCl2 as a limiting reagent
mol AgCl :
= 2/1 x 0.161
= 0.322
Mass AgCl :
= 0.322 x 143,32 g/mol
= 46.149 g
mol remains(unreacted) for AgNO3 :
= 0.344-(2/1 x 0.161)
= 0.022
mass AgNO3 remains :
= 0.022 x 169,87 g/mol
= 3.74 g
The density of the unknown metal, given the data is 1.67 g/mL
<h3>What is density? </h3>
The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as
Density = mass / volume
With the above formula, we can determine the density of the unknown metal. Details below:
<h3>How to determine the density </h3>
- Mass of unknown metal = 20 - 0 = 20 g
- Volume of unknown metal = 12 - 0 = 12 mL
- Density of unknown metal =?
Density = mass / volume
Density of unknown metal = 20 / 12
Density of unknown metal = 1.67 g/mL
Thus, the density of unknown metal is 1.67 g/mL
Learn more about density:
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Answer: option C, a mixture
Explanation:
Crude oil is a mixture of hydrocarbons and different organic materials. It is refined into useful materials like petrol,diesel and various petrochemicals.
2C6H14 + 19O2 ===> 12CO2 + 14H2O ... balanced equation
moled hexane present = 2.6 g x 1 mole/130 g = 0.02 moles
moles O2 present = 5.29 g x 1 mole/32 g = 0.165 moles
Which reactant is limiting? Hexane = 0.02/2 = 0.01; O2 = 0.165/19 = 0.0087
Thus O2 is limiting...
moles of H2O that can be produced =0.165 moles O2 x 14 H2O/19 CO2 = 0.122 moles H2O
Mass H2O = 0.122 moles x 18 gm/mole = 2.20 g (to 3 sig. figs.)
