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natali 33 [55]
3 years ago
8

For the reaction 2Fe + O2 = 2FeO, how many grams of iron oxide are produced from 8.00 mol of iron? when o2 is an excess

Chemistry
1 answer:
Yanka [14]3 years ago
6 0

Answer:

The answer to your question is 576 grams of FeO

Explanation:

Balanced chemical reaction

                      2Fe  +  O₂  ⇒  2 FeO

       Reactant             Elements         Products

             2                         Fe                     2

             2                         O                       2

Data

mass of FeO = ?

moles of Fe = 8

Excess Oxygen

Process

1.- Convert moles of Fe to grams

Atomic mass = 56 g

                         56 g of Fe ------------------ 1 mol

                           x               ------------------ 8 moles

                           x = (8 x 56) / 1

                           x = 448 g of Fe

2.- Use the coefficients of the balanced reaction to find the grams of FeO.

Molecular mass of FeO = 2[ 56 + 16]

                                       = 144 g

               2(56) g of Fe ----------------------- 144 g of FeO

               448 g             -----------------------  x

                x = (448 x 144) / 2(56)

                x = 64512 / 112

               x = 576 grams of FeO

                       

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The balanced chemical reaction will be:

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0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

<h3>What is baking powder?</h3>
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  • The addition of a buffer, such as cornstarch, prevents the base and acid from reacting prematurely.
  • Baking powder is used in baked goods to increase volume and lighten the texture.

To find how many moles of CO₂ are produced from 1.00 g of baking powder:

The balanced equation is:

  • Ca(H₂PO₄)₂(s) + 2NaHCO₃(s) → 2CO₂(g) + 2H₂O(g) + CaHPO₄(s) + Na₂HPO₄(s)

On 3.50 g of baking power:

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The molar masses are: Ca = 40 g/mol; H = 1 g/mol; P = 31 g/mol; O = 16 g/mol; Na = 23 g/mol; C = 12 g/mol.

So,

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The number of moles is the mass divided by molar mass, so:

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  • nNaHCO₃ = 1.085/84 = 0.0129 mol

First, let's find which reactant is limiting.

Testing for Ca(H₂PO₄)₂, the stoichiometry is:

  • 1 mol of Ca(H₂PO₄)₂ ---------- 2 mol of NaHCO₃
  • 0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

  • x = 0.012 mol

So, NaHCO₃ is in excess.

The stoichiometry calculus must be done with the limiting reactant, then:

  • 1 mol of Ca(H₂PO₄)₂ ------------- 2 mol of CO₂
  • 0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

  • x = 0.012 mol of CO₂

Therefore, 0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

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The correct question is given below:

Calcium dihydrogen phosphate, Ca(H2PO4)2, and sodium hydrogen carbonate, NaHCO3, are ingredients of baking powder that react with each other to produce CO2, which causes dough or batter to rise: Ca(H2PO4)2(s) + NaHCO3(s) → CO2(g) + H2O(g) + CaHPO4(s) + Na2HPO4(s)[unbalanced] If the baking powder contains 31.0% NaHCO3 and 35.0% Ca(H2PO4)2 by mass: (a) How many moles of CO2 are produced from 3.50 g of baking powder?

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Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

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The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

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1 mole of F₂ reacts with 2 moles of NaCl

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1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

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The mass of the NaCl ≈ 132.6 gams

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The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

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The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

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