Question

Please help !!

Molecules CaSiO3
2. Before conducting a chemical reaction, I complete my RICE table and find that I should get
84.9 g of my desired product. However, after running the experiment, I actually end up with 78.4
g of my desired product. What was my percent yield?
% yield
3.
Balanced chemical reaction: Ca(CO3) -
3 CO,+ CaO
Based on the above balanced
chemical reaction, how many Liters of CO2 gas (at STP)
should be formed if the mass of CO2 is 24 grams?
L of CO2 gas
4. I burned 280 grams of CH4O2 gas with 7.2 moles of Oxygen gas, according to the following
67 words

Answer 1

For 2: The % yield of the product is 92.34 %

For 3: 12.208 L of carbon dioxide will be formed.

Step-by-step explanation:

• For 2:

The percent yield of a reaction is calculated by using an equation:

`\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}* 100` ......(1)

Given values:

Actual value of the product = 78.4 g

Theoretical value of the product = 84.9 g

Plugging values in equation 1:

`\% \text{yield}=(78.4 g)/(84.9g)* 100\\\\\% \text{yield}=92.34\%`

Hence, the % yield of the product is 92.34 %

• For 3:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` ......(2)

Given mass of carbon dioxide = 24 g

Molar mass of carbon dioxide = 44 g/mol

Plugging values in equation 1:

`\text{Moles of carbon dioxide}=(24g)/(44g/mol)=0.545 mol`

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

So, 0.545 moles of carbon dioxide will occupy =
`(22.4L)/(1mol)* 0.545mol=12.208L` of volume

Hence, 12.208 L of carbon dioxide will be formed.