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KiRa [710]
3 years ago
6

How many moles of oxygen are required to react with methanol in order to produce 13 moles of formaldehyde?

Chemistry
1 answer:
ddd [48]3 years ago
3 0

Answer:

6.5 moles of Oxygen are required

Explanation:

Based on the reaction:

CH3OH + 1/2 O2 → CH2O + H2O

1 mole of methanol reacts with 1/2 moles O2 to produce 1 mole of formaldehyde and 1 mole of water.

Thus, to produe 13 moles of formaldehyde, CH2O, are needed:

13 moles CH2O * (1/2mol O2 / 1mol CH2O) =

<h3>6.5 moles of Oxygen are required</h3>
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An oxybromate compound, kbrox, where x is unknown, is analyzed and found to contain 43.66 % br.
aniked [119]

Answer is: an oxybromate compound is KBrO₄ (x = 4).

ω(Br) = 43.66% ÷ 100%.

ω(Br) = 0.4366; mass percentage of bromine.

If we take 100 grams of compound:

m(Br) = ω(Br) · 100 g.

m(Br) = 0.4366 · 100 g.

m(Br) = 43.66 g; mass of bromine.

n(Br) = m(Br) ÷ M(Br).

n(Br) = 43.66 g ÷ 79.9 g/mol,

n(Br) = 0.55 mol; amoun of bromine.

From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).

m(K) = 0.55 mol · 39.1 g/mol.

m(K) = 21.365 g; mass of potassium in the compound.

m(O) = 100 g - 21.365 g - 43.66 g.

m(O) =34.97 g; mass of oxygen.

n(O) = 34.97 g ÷ 16 g/mol.

n(O) = 2.185 mol.

n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.

n(K) : n(Br) : n(O) = 1 : 1 : 4.

6 0
3 years ago
A sample of 211 g of iron (III) bromide is reacted with
Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

  • FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

  • Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

\tt n=\dfrac{186}{78.0452}=2.38

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

\tt \dfrac{6}{3}\times 0.714=1.428

6 0
3 years ago
16.
mixas84 [53]

Answer:

Explanation:

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8 0
2 years ago
What is 30% NaOH test​
pogonyaev

Sodium hydroxide solution 30% CAS 1310-73- 2 Suprapur® - Find MSDS or SDS, a COA, data sheets and more information.

CAS number:

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3 years ago
Epsom salts, a strong laxative used in veterinary medicine, is a hydrate, which means that a certain number of water molecules a
alexira [117]
The values of x represents that number of moles of water molecules that is present per mole of the salt magnesium sulfate. To determine the value for this, we need to know how much is the water that is lost after heating the sample assuming that all of the water molecules are evaporated leaving only the unhydrated form of the salt. We calculate as follows:

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The mass lost is equal to the mass of water lost.

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moles water / moles MgSO4 = 0.0989 mol H2O / 0.0141 mol MgSO4 = 7

Therefore, the value of x is 7.
8 0
3 years ago
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