Mass of PH3= 6.086 g
<h3>Further explanation</h3>
Given
6.0 L of H2
Required
mass of PH3
Solution
Reaction
P4 + 6H2 → 4PH3
Assumed at STP ( 1 mol gas=22.4 L)
Mol of H2 for 6 L :
= 6 : 22.4 L
= 0.268
From the equation, mol PH3 :
= 4/6 x moles H2
= 4/6 x 0.268
= 0.179
Mass PH3 :
= 0.179 x 33,99758 g/mol
= 6.086 g
There are:
3.41 moles of C
4.54 moles of H
3.40 moles of O.
Why?
To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.
We have that:

To know the percent of each element, we need to to the following:

So, we know that for the 100 grams of the compound, there are:
40.92 grams of C
4.58 grams of H
54.49 grams of O
We know the molecular masses of each element:

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.
Have a nice day!
Describe the process by which Ag+ ions are precipitated out of solution. 4. In your testing, several precipitates are formed, and then dissolved as complexes.