The actual yield is 43 g Cl₂.
The <em>limiting reactant was MnO₂</em> because it gave the smaller mass of Cl₂.
∴ The <em>theoretical yield</em> is 60.25 g Cl₂.
% yield = actual yield/theoretical yield × 100 %
Actual yield = theoretical yield × (% yield/100 %) = 60.25 g × (72 %/100%) = 43 g
Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.
First, how many moles of each substance are there
the molar mass of BCl3 is <span>117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.
Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles.
But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.
.96 x 36.46 = ~35 g</span>
Explanation:
dredge to do with me to do with me to do with me to do with me to do with me to
Answer:
Explanation:
We are given two reactions which are the two steps of a mechanism:
......(1)
.......(2)
To determine the net chemical equation, we will multiply equation 2
.......(3)
Adding (1) and (3)
Thus, the net balanced chemical equation is:
I am going to have to say C. luster if not then B. Cleavage