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Ivahew [28]
3 years ago
12

How many moles of oxygen gas are required to form 55g of carbon monoxide gas

Chemistry
1 answer:
marta [7]3 years ago
6 0

Answer:

8.33 moles CO2 X. 25mol O2. 16mol CO2. = 13.0 moles

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Describe two physical and two chemical changes involved in cooking. (4 points)
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Physical changes are when things get changed without altering chemical consistencies, which is melting solid butter into liquid one, or boiling water. Chemical changes are things such as caramelizing sugar when making sweets, or when carbon dioxide is created and released when baking bread.
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Which is the largest?<br> O 43.32 mL<br> 0 0,0056 L<br> O 0.000076 KL
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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 <u>reaction</u>:

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 <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:

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 <u>molar mass</u> 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 <u>molar ratio</u> 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 <u>grams of magnesium carbonate</u>:

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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3 years ago
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iris [78.8K]

Answer:

I'd Go. B. weigh everything, let the reaction happen, then weigh everything again.

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