00:01
In this problem, we are given a chemical reaction and told that 800 milliliters of 0 .5 molar nitric acid is added to 3 .177 of copper.
00:10
After the reaction, we find that there is 0 .0739 grams of copper remaining.
00:17
We're asked to find the extent of reaction and then calculate the moles of nitrogen monoxide formed and the moles of nitric acid remaining.
00:26
To start solving for the extent of reaction, we can relate our stoichiometric coefficients and what's being consumed or formed to the extent of reaction which is represented here.
00:38
So the stochometric coefficient is given a negative sign when it's involved with reactants and a positive sign that's involved with products.
00:47
We can solve for our starting amount of moles of copper by taking our initial mass and dividing by the molar mass of copper.
01:00
So we find that initially we had 0 .05 moles of copper.
01:07
To calculate the moles of copper that we have remaining, we can take take the final amount of grams that we are told we have, and again, divide by the molar mass.
01:22
So we find that at the end of our reaction, we have 0 .00 -116 moles of copper remaining.
01:31
We can set up a relationship to solve for our extent of reaction.
01:35
So we know that we start with 0 .05 moles of copper.
01:41
And during the reaction, we consume 3 times our extent of reaction...