Answer: 31.8 g
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require 3 moles of 
Thus 0.59 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent as it is present in more amount than required.
As 1 mole of give = 2 moles of 
Thus 0.59 moles of give =
of
Mass of 
Thus 31.8 g of will be produced from the given masses of both reactants.
Answer: D is right
Explanation: One mole contains 6.0225 ·10^23 molecules
Despite of the substance
Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
Follow
these steps to solve the given equation:
Multiply
the two decimal figures together and find the sum of the exponents, that is,
(1.5
* 1.89) * 10 ^4+3
(2.835)
* 10^7
10^7
can also be written as e.70
'e'
stands for exponential.
Therefore,
we have 2. 835 e 7.0 = 2.8 e 7.0.
Based on the calculations above, the correct option is A.
