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alekssr [168]
3 years ago
10

How many grams of carbon dioxide are produced from the combustion of 1.3 moles of acetylene

Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0
Complete combustion of acetylene generated carbon dioxide and water. This can be represent by following reaction

2C2H2   +    5O2     →      4CO2     +     2H2O
(2 mole)                            (4 mole)

From the above balanced reaction, it can be seen that 2 mole of acetylene on complete combustion generates 4 moles of carbon dioxide
i.e. 2 mole of C2H2 ≡ 4 mole of CO2
∴ 1.3 mole of C2H2 ≡ (4 X 1.3)/2 = 2.6 mole of CO2

Now, 1 mole of CO2 = 44 g
∴ 2.6 mole of CO2 = (44 X 2.6) = 114.4 g

Thus,  <span>114.4 grams of carbon dioxide are produced from the combustion of 1.3 moles of acetylene</span>
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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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