Answer:
We need 1.714 moles N2
Explanation:
Step 1: Data given
The reaction yield = 87.5 %
Number of moles NH3 produced = 3.00 moles
Step 2: The balanced equation
N2(g)+ 3H2(g) →2NH3(g)
Step 3: Calculate moles N2
For 2 moles NH3 produced we need 1 mol N2 and 3 moles H2
This means, if the yield was 100%, for 3.00 moles NH3 produced , we need 1.5 moles N2
For a 87.5 % yield:
we need more N2, increased by a ratio of 100/87.5.
100/87.5 * 1.5 = 1.714 moles N2
Answer: It has to be 56.10564
Explanation:1 grams KOH to mol = 0.01782 mol
10 grams KOH to mol = 0.17824 mol
20 grams KOH to mol = 0.35647 mol
30 grams KOH to mol = 0.53471 mol
40 grams KOH to mol = 0.71294 mol
50 grams KOH to mol = 0.89118 mol
100 grams KOH to mol = 1.78235 mol
200 grams KOH to mol = 3.5647
Answer:
<h2>134km = 13400000cm</h2><h2>35g = 35000000ug</h2><h2>0.65mmol = 0.00065mol</h2>
It is C 2 1 2 1
You have to 1st balance the nitrates then balance the silvers to get the coefficients
