A sample of Xe takes 75 seconds to effuse out of a container. An unknown gas takes 37 seconds to effuse out of the identical con
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Given equation :
Given information = 2.47 grams KNO2 and excess KMnO4 and we need to find grams of water (H2O).
Since KMnO4 is in excess, so grams of water(H2O) can be calculated using grams of KNO2 with the help of stoichiometry.
To find grams of water(H2O) from grams of KNO2 , we need to follow three steps.
Step 1. Convert 2.47 grams of KNO2 to moles of KNO2.
Molar mass of KNO2 = 85.10 g/mol
Moles = 0.0290 mol KNO2
Step 2. Convert moles of KNO2 to moles of H2O using mole ratio.
Mole ratio are the coefficient present in front of the compound in the balanced equation.
Mole ratio of KNO2 : H2O is 5 : 3 (5 coefficient of KNO2 and 3 coefficient of H2O)
Mole = 0.0174 mol H2O
Step 3. Convert mole of H2O to grams of H2O
Grams = Moles X molar mass
Molar mass of H2O = 18.00 g/mol
Grams of water = 0.313 grams H2O
Summary : The above three steps can also be done in a singe setup as shown below.
In the above setup similar units get cancelled out and we will get grams of H2O as 0.313 grams water (H2O)
Considering the definition of percentage by mass, the mass percentage of CaCO₃ is 68.59%.
<h3>What is mass percentage</h3>The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.
In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.
The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:
In this case, you know:
- mass of CaCO₃: 2.62 grams
- mass of limestone: 3.82 grams
Replacing in the definition of mass percentage:
<u><em>mass percentage= 68.59 %</em></u>
Finally, the mass percentage of CaCO₃ is 68.59%.
Learn more about percentage by mass:
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