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Harrizon [31]
1 year ago
13

A 13.00 g sample of citric acid reacts with an excess of baking soda as shown in the equation.

Chemistry
2 answers:
UNO [17]1 year ago
8 0
From Law of definite proportion
You can compare this
Citric acid (Mw) / 3CO2 (Mw) = Citric acid (g) / 3CO2 (g)
So (12)6+(1)8+(16)7 / 3[12 + (16)2] = 13 / 3CO2
192 / 132 = 13 / CO2
You can get 3CO2 = 8.93 g
PolarNik [594]1 year ago
4 0

Answer:

eheheehehehszndn!jejxxnndrrjrrrfufurururufjththjrjrjdjjjrj\u\ujrjeejrjjjj carbon

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Displacement is the distance and direction of an object's change in position from the starting.
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The chemical formula of lead sulfide tells you that it contains<br>​
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How many atoms are in 1.75 mol CHCL3
vaieri [72.5K]

1 mol = 6.022 x 10²³ atoms

In order to find how many atoms, dimly multiply the amount of moles you have by 6.022 x 10²³ or Avogadro's number.

So you have 1.75 mol CHC1₃ x (6.022x10²³) = 1.05385 x 10²⁴ atoms of CHCl₃

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6 0
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What does the sun do to generate energy?
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5 0
3 years ago
Read 2 more answers
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
3 years ago
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