Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, 
.
Explanation:
Given: Mass of methane = 146.6 g
As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.
The given reaction equation is as follows.
This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.
Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, .
An intensive property does not change when you take change when you take away some of the sample the producers that a student could use to examine the intensive property of a rectangular block of wood
Answer:
this reaction is an oxidation reaction
Note that it says oxygen "gas"
So you need the atomic mass of oxygen gas
Look at your periodic table, you'll see 15.9994 under oxygen
Oxygen gas has a formula of O2 therefore,
(15.9994) times 2= Oxygen gas atomic mass=31.9988
Mol= Mass/Atomic Mass
=62.3 g/ 31.9988 g/mol = 1.95 mol
now look at the ratio of C2H6 and O2, notice there is an invisible number beside each of them, at that "invisible number" is =1
1 C2H6 + 1 O2 -> products
this means that for 1 mol of C2H6, 1 mol of O2 has to react with it
Thus as we have 1.95 moles of O2, we need 1.95 moles of C2H6
The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg
Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol
The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺
The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5
Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
= 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol
Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
= 0.031395 g
= 31.395 mg
≈ 31.4 mg
