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3 years ago

2. A lab group turned in their experiment and had measured the mass of their sample

Chemistry

1 answer:

alisha [4.7K]3 years ago

3 0

Answer:

1.61%.

Explanation:

The following data were obtained from the question:

Measured mass = 3.15 g.

Actual mass = 3.10 g

Percentage error =.?

The percentage error of the group can be obtained by using the following equation below:

Percentage error = |Actual mass – Measured mass| / Actual mass × 100

Percentage error = |3.10 – 3.15| /3.10 × 100

Percentage error = 0.05/3.1 × 100

Percentage error = 5/3.1

Percentage error = 1.61%

Therefore, the percentage error of the group is 1.61%.

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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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