Answer: 16.32 g of as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of
Thus 16.32 g of as excess reagent are left.
Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:
In terms of mass, specific heat and temperature change is:
Now, solve for the final temperature, as follows:
Then, plug in the masses, specific heat and temperatures to obtain:
Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.
Regards!
