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Whitepunk [10]
3 years ago
11

iron will react with oxygen to produce Fe2O3. how many grams of Fe2O3 will he produced if 0.18 mil of Fe reacts?

Chemistry
1 answer:
nasty-shy [4]3 years ago
6 0

Answer:

14.4g

Explanation:

First, we need to write a balanced equation for the reaction between Fe and O2 to produce Fe2O3. This is illustrated below:

4Fe + 3O2 —> 2Fe2O3

From the balanced equation,

4moles of Fe produced 2moles of Fe2O3.

Therefore, 0.18mol of Fe will produce = (0.18x2) /4 = 0.09mol of Fe2O3.

Now we need to find the mass present in 0.09mol of Fe2O3. This can be achieved by doing the following:

Molar Mass of Fe2O3 = (56x2) + (16x3) = 112 + 48 = 160g/mol

Number of mole of Fe2O3 = 0.09mol

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Fe2O3 = 0.09 x 160 = 14.4g

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5000

Explanation:

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3 years ago
What volume of carbon dioxide (CO2) will be produced if 2.90 moles of it on (Fe) is produced?
dolphi86 [110]

Answer:

= 97.44 Liters at S.T.P

Explanation:

The reaction between Iron (iii) oxide and Carbon monoxide is given by the equation;

Fe2O3(s)+ 3CO(g) → 3CO2(g) + 2Fe(s)

From the reaction when the reactants react, 2 moles of Fe and 3 moles of CO2 are produced.

Therefore; Mole ratio of Iron : Carbon dioxide is 2:3

Thus; Moles of Carbon dioxide = (2.9/2)×3

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But; 1 mole of CO2 at s.t.p occupies 22.4 liters

Therefore;

Mass of CO2 = 22.4 × 4.35 Moles

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6 0
3 years ago
For an aqueous solution of sodium chloride (NaCl) .Determine the molarity of 3.45L of a solution that contains 145g of sodium
worty [1.4K]
Molar mass NaCl = 58.44 g/mol

number of moles:

mass NaCl / molar mass

145 / 58.44 => 2.4811 moles of NaCl

Volume = 3.45 L

Therefore :

M = moles / volume in liters:

M = 2.4811 / 3.45

M = 0.719 mol/L⁻¹

hope this helps!
6 0
3 years ago
What is the molarity of a solution prepared by dissolving 2.5 grams of LiNO3 in sufficient water to make 60.0 mL of solution?​
Murrr4er [49]

Answer: hope this helps

To make molar NaCl solutions of other concentrations dilute the mass of salt to 1000ml of solution as follows:

0.1M NaCl solution requires 0.1 x 58.44 g of NaCl = 5.844g.

0.5M NaCl solution requires 0.5 x 58.44 g of NaCl = 29.22g.

2M NaCl solution requires 2.0 x 58.44 g of NaCl = 116.88g.

Explanation:

6 0
3 years ago
Read 2 more answers
For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temper
Keith_Richards [23]

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1\text{ and }V_1 are initial pressure and volume at initial temperature T_1.

P_2\text{ and }V_2 are final pressure and volume at initial temperature T_2.

We are given:

P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V

T_1=-25^oC=248.15 K

T _2 = 16^oC=289.15 K

Putting values in above equation, we get:

\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}

P_1=0.225 kPa

Hence, the new pressure will be 0.225  kPa.

4 0
3 years ago
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