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3 years ago

If 9.5 × 10²⁵ molecules of CO₂ are produced in a combustion reaction, what is the mass in kg of CO₂ that is produced?

Chemistry

1 answer:

Leno4ka [110]3 years ago

4 0

Answer:

6.9428 kg of CO2

Explanation:

1) Use Avogadro's number that states 1 mole = 6.022 x 10^23 particles. Convert 9.5 x 10^25 molecules into moles.

9.5 x 10^25 CO2 molecules $x\frac{1 mole CO2}{6.022 x 10^23}$ = 157.75 moles CO2

2) Convert 157.75 moles of CO2 into grams. CO2's molar mass is 44.01g.

$157.75moles CO2$ $x \frac{44.01g/mol}{1 mole CO2}$ = 6942.79g

3) Convert 6942.79 grams into kilograms (divide by 1000):

6.9428 kg

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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!

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boyakko [2]

Answer:

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Explanation:

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Nickel and carbon monoxide react to form nickel carbonyl, like this: (s)(g)(g) At a certain temperature, a chemist finds that a

horsena [70]

The question is incomplete, here is the complete question:

Nickel and carbon monoxide react to form nickel carbonyl, like this:

$Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)$

At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:

Compound Amount

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