The percent yield of this reaction is calculated as follows
Mg3N2 + 3H2O =2NH3 + 3Mgo
calculate the theoretical yield,
moles=mass/molar mass
moles Mg3N2= 3.82 g/100g/mol= 0.0382 moles(limiting regent)
moles of H2o= 7.73g/18g/mol = 0.429 moles ( in excess_)
by use of mole ratio between Mg3N2 to MgO which is 1:3 the moles of MgO = 0.0382 x3 = 0.1146 moles
mass =moles x molar mass
the theoretical mass is therefore = 0.1146mole x 40 g/mol = 4.58 grams
The % yield = actual mass/theoretical mass x1000
= 3.60/4.584 x100= 78.5%
The answer is either A or B I’m not sure
The answer for the following problem is mentioned below.
- <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.</em></u>
Explanation:
Given:
mass of calcium phosphate (
) = 125.3 grams
We know;
molar mass of calcium phosphate (
) = (40×3) + 3 (31 +(4×16))
molar mass of calcium phosphate (
) = 120 + 3(95)
molar mass of calcium phosphate (
) = 120 +285 = 405 grams
<em>We also know;</em>
No of molecules at STP conditions(
) = 6.023 × 10^23 molecules
To solve:
no of molecules present in the sample(N)
We know;
N÷
=
N =(405×6.023 × 10^23) ÷ 125.3
N = 19.3 × 10^23 molecules
<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules</em></u>
Answer:
1s2 2s2 2p2
Explanation:
it has 6 electrons in two energy levels so the sub levels are 1s, 2s and 2p