For the following: a)give the reaction type; b)complete and balance the reaction; c)using your answer to part b, determine how much of the other reactant is needed, and how much of each product will be formed, for the given starting amount (below); and d) calculate the percent yield
potassium metal reacts with aqueous calcium nitrate yielding ...
*start with 101.3467 mol of potassium
*if 20 mol of Ca are collected
This is a lot of work and you need to know how to do this type problem. So tell us what you understand, how much you can do, and what you don't understand and we can help you through it. I'm not interested in spending the next hour typing the answers just so you can check your work. But I'll be happy to check your work if you type your answers.
To answer these questions, we need to follow the steps below:
a) Give the reaction type:
To determine the reaction type, we need to write out the chemical equation for the reaction. The given information states that potassium metal reacts with aqueous calcium nitrate. So the chemical equation is:
2K + Ca(NO3)2 → 2KNO3 + Ca
The reaction type for this equation is a double displacement reaction.
b) Complete and balance the reaction:
To balance the reaction, we need to ensure that the number of atoms of each element is equal on both sides of the equation. Taking a look at the equation, we see that there is one calcium (Ca) atom on the right side, while there are two potassium (K) atoms on the left side. Therefore, we need to balance the equation by placing a coefficient of 2 in front of the potassium:
2K + Ca(NO3)2 → 2KNO3 + Ca
Now the equation is balanced.
c) Determine the amount of the other reactant needed and the amount of each product formed:
From the balanced chemical equation, we can see that 2 moles of potassium reacts with 1 mole of calcium nitrate, producing 2 moles of potassium nitrate and 1 mole of calcium.
Given that the starting amount of potassium is 101.3467 mol and 20 mol of calcium are collected, we can use stoichiometry to find the amounts of calcium nitrate and potassium nitrate formed.
From the balanced equation, the molar ratio between potassium and calcium nitrate is 2:1, and the molar ratio between potassium and potassium nitrate is 2:2 (or simply 1:1).
101.3467 mol of potassium is enough to react with (101.3467/2) = 50.67335 mol of calcium nitrate.
Since the ratio of potassium to potassium nitrate is 1:1, this means that (101.3467 mol) of potassium will produce the same amount of potassium nitrate.
Therefore, 101.3467 mol of potassium will react to produce 50.67335 mol of calcium nitrate and 101.3467 mol of potassium nitrate.
d) Calculate the percent yield:
To calculate the percent yield, we need the actual yield and the theoretical yield. The actual yield is given as 20 mol of calcium.
The theoretical yield can be calculated using the balanced equation. From the equation, we can see that the stoichiometric ratio between calcium and potassium is 1:1. Therefore, for every 1 mole of calcium collected, 2 moles of potassium react.
Since 20 mol of calcium were collected, the theoretical yield of potassium would be (20 mol × 2) = 40 mol.
To calculate the percent yield, we use the formula:
Percent Yield = (Actual Yield / Theoretical Yield) × 100%
Percent Yield = (20 mol / 40 mol) × 100%
Percent Yield = 50%
Therefore, the percent yield is 50%.