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2 years ago

How many liters of oxygen are required to react completely with 14.8 mol of Al?

Chemistry

1 answer:

Bond [772]2 years ago

8 0

Answer:

296 L

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

4Al + 3O₂ ⟶ 2Al₂O₃

n/mol: 17.4

1. Moles of O₂

$n = \text{17.4 mol Al}\times \dfrac{\text{3 mol O}_{2}}{\text{4 mol Al}}= \text{13.05 mol O}_{2}$

2. Volume of O₂

You haven't given the conditions at which the volume is measured, so I assume it is at STP (0 °C and 1 bar).

At STP the molar volume of a gas is 22.71 L.

$V = \text{13.05 mol}\times \dfrac{\text{22.71 L}}{\text{1 mol }}= \textbf{296 L}$

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How many moles of CaO will be produced from 23.9 g of Ca? 2Ca(s) + O2(g) – 20:0(3) mol

inessss [21]

Answer:

0.6 moles of CaO will produced.

Explanation:

Given data:

Mass of calcium = 23.9 g

Moles of CaO produced = ?

Solution:

Chemical equation:

2Ca + O₂ → 2CaO

Number of moles of calcium:

Number of moles = mass/ molar mass

Number of moles = 23.9 g / 40 g/mol

Number of moles = 0.6 mol

Now we will compare the moles of calcium and CaO.

Ca : CaO

2 : 2

0.6 : 0.6

0.6 moles of CaO will produced.

4 0

2 years ago

Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following

bearhunter [10]

Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the reaction:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the number of moles that we have in the 110.0 g of carbon dioxide, to this, we have to know the atomic mass of each atom:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the molar mass of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$

In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the molar ratio between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the grams of magnesium carbonate:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

I hope it helps!

8 0

2 years ago

Please help its easy but i don't feel like doing it

KonstantinChe [14]

Answer:

first one is a second one is e

Explanation:

7 0

3 years ago

Complete combustion of a sample of a hydrocarbon on excess oxygen produces equimolar quantities of carbon dioxide and water. Whi

Bond [772]

C. C4H8 Given that the number of moles of CO2 and H2O produced from the combustion are equal, that means for every carbon atom, there are 2 hydrogen atoms because CO2 has only 1 carbon atom and H2O has 2 hydrogen atoms. So let's look at the available choices and see which one is correct. A. C2H2 This is a 1 to 1 ratio of carbon to hydrogen. Wrong answer. B. C2H6 This is a 1 to 3 ratio of carbon to hydrogen. Wrong answer. C. C4H8 This is a 1 to 2 ratio of carbon to hydrogen. Correct answer. D. C6H6 This is a 1 to 1 ratio of carbon to hydrogen. Wrong answer.

6 0

2 years ago

Does anybody know how to do this if so please help !

Viktor [21]

Answer:

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Explanation:

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3 0

2 years ago