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3 years ago

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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F a sample of butene (C4H8) that has a mass of 136.6 g is combusted in excess oxygen, what is the mass of CO2 that is produced?

alukav5142 [94]

The balanced chemical reaction will be:

C4H8 + 6 O2 --> 4 CO2 + 4 H2O

We are given the amount of butene being combusted. This will be our starting point.

136.6 g C4H8 (1 mol C4H8/ 56.11 g C4H8) (4 mol CO2/1 mol <span>C4H8</span>) ( 44.01 g CO2/ 1 mol CO2) = 428.6 g CO2

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3 years ago

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Calcium dihydrogen phosphate, Ca(H₂PO₄)₂, and sodium hydrogen carbonate, NaHCO₃, are ingredients of baking powder that react to

NikAS [45]

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

<h3>What is baking powder?</h3>

Baking powder is a dry chemical leavener composed of carbonate or bicarbonate and a weak acid.
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:

mCa(H₂PO₄)₂ = 0.35 × 3.50 = 1.225 g
mNaHCO₃ = 0.31 × 3.50 = 1.085 g

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,

Ca(H₂PO₄)₂: 40 + 4 × 1 + 31 + 8 × 16 = 203 g/mol
NaHCO₃: 23 + 1 + 12 + 3 × 16 = 84 g/mol

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

nCa(H₂PO₄)₂ = 1.225/203 = 0.006 mol
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.

Know more about baking powder here:

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#SPJ4

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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1 year ago

What is the name of the compound (NH4)2SO4

GREYUIT [131]

The answer is Ammonium sulfate

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3 years ago

Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat

Alika [10]

Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

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

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

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

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

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

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

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

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

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

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

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

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3 years ago

Which is NOT an example of a solution?

USPshnik [31]

C table snap Lt it is a compound

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3 years ago

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