What is the average atomic mass of sodium

how many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 2.75 g of fe2s3 if the percent yield

Snezhnost [94]

Answer:

We need 203 mL of FeCl3

Explanation:

Step 1: Data given

molarity FeCl3 = 0.200 M

Mass of Fe2S3 = 2.75 grams

Percent yield = 65.0 %

Step 2: The balanced equation

3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)

Step 3: Calculate moles Fe2S3

Moles Fe2S3 = mass Fe2S3 / molar mass Fe2S3

Moles Fe2S3 = 2.75 grams / 207.9 g/mol

Moles Fe2S3 = 0.0132 moles

Step 4: Calculate theoretical yield

65.0 % = 0.65 = actual yield/ theoretical yield

theoretical yield = actual yield / 0.65

theoretical yield = 0.0132 moles /0.65

theoretical yield = 0.0203 moles Fe2S3

Step 5: Calculate moles FeCl3

For 1 mol Fe2S3 and 6 mol NaCl we need 3 moles Na2S and 2 moles FeCl3

For 0.0203 moles Fe2S3 we need 2*0.0203 = 0.0406 moles FeCl3

Step 6: Calculate volume

Molarity = moles / volume

Volume = moles / molarity

Volume = 0.0406 moles / 0.200 M

Volume = 0.203 L = 203 mL

We need 203 mL of FeCl3

6 0

4 years ago

I need help with this question I don’t understand it at all!!!

bezimeni [28]

Answer:

C-14 => N-14 + β⁻

Explanation:

C-14 over time decays by beta emission to N-14. Before decay, C-14 nucleus contains 6 protons and 8 neutrons. It is accepted that a neutron (n°) is composed of 1 proton (p⁺) and 1 electron (e⁻). Beta emission indicates loss/discharge of a high energy e⁻ from the nucleus of C-14 leaving 7 protons and 7 neutrons. The number of protons defines the element, N-14 as a product of the decay process with the β particle (high energy electron) the other. To check, remember, the ∑mass Reactants = ∑mass Products (superscript numbers) and ∑reactant charges = ∑product charges (subscript numbers

₆C¹⁴ => ₇N¹⁴ + ₋₁e° or, C-14 => N-14 + β

4 0

3 years ago

what mass of Fe2O3 is produced in the reaction in the table above mass of reactants= 223.4 g Fe+96.0 g O2. and mass of products=

Ganezh [65]

Answer:

AS ACCORDING TO THE LAW OF MASS CONSERVATION

REACTANTS =PRODUCTS

THEREFORE,

223.4+96=MASS OF FE2O3

=319.4 FE203

4 0

3 years ago

2 C6H14 + 19 O2 --> 12 CO2 + 14 H2O

Rashid [163]

Answer:

4.06 mol H₂O

Explanation:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

First we convert the given masses of reactants into moles, using their respective molar masses:

250 g O₂ ÷ 32 g/mol = 7.81 mol O₂

50 g C₆H₁₄ ÷ 86 g/mol = 0.58 mol C₆H₁₄

Now we calculate how many O₂ moles would react completely with 0.58 C₆H₁₄ moles, using the stoichiometric coefficients of the reaction:

0.58 mol C₆H₁₄ * $\frac{19molO_2}{2molC_6H_{14}}$ = 5.51 mol O₂

As there are more O₂ moles than required (7.81 vs 5.51), O₂ is the reactant in excess. That means that C₆H₁₄ is the limiting reactant.

Now we can calculate how much water can be formed, using the number of moles of the limiting reactant:

0.58 mol C₆H₁₄ * $\frac{14molH_2O}{2molC_6H_{14}}$ = 4.06 mol H₂O

5 0

3 years ago

Help me for this. Thankyou!

Arte-miy333 [17]

Answer:

I cant really see the worksheet can u try again

Explanation:

7 0

3 years ago

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