Use the question marck Moles of CO2
The the giving = 0.624 mol O2
Find the CF faction = 1 mole= 32.00 of O2
O= 2x16.00= 32.00amu ( writte this in the cf fraction)
SET UP THE CHART
Always start with the giving
0.624 mol O2 / 1mol of CO2
___________ / _____________ = Cancel the queal ( O2)
/ 32.00c O2
/
/
Multiply the top and divide by the bottom
0.624 mol CO x 1mol CO2 = 0.624 divide by 32.00 O2 =0.0195
You should look at the giving number ( how many num u gor ever there)
Ur answer should have the same # as ur givin so
= 0.0195
= .0195 mol of CO2
<span>We look at the end of the day:
n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
Answer:
A. 
B. 
Explanation:
Hello!
In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:
Thus we proceed as follows:
A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:
Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:
B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:
Best regards!
<span>H</span>₃<span>PO</span>₄<span> - Phosphoric Acid :
H</span>₃<span>PO</span>₄<span> = 1.0 * 3 + 31.0 + 16.0 * 4 = 98.0 g/mol
hope this helps!</span>
