ph3 occupied 45.4L at STP calculate mass of PH and number of particle
Answers
Answer:
Explanation:
Solution :
[1 mole of any substances contains 6.023×1023 number of molecules [Avogadro's number]
(a) 6.022×1023 molecules occupied 22.42 cm3 (its) =22.4×6.023×10226.023×1023 = 2.24 lits.
Answer:
easy I hope it helps you
Explanation:
Approximately
12.4
grams of phosphine will be produced.
Explanation:
Start by writing a balanced chemical equation. Here is the unbalanced:
P
4
+
H
2
→
P
H
3
Comparing coefficients to balance, we get:
P
4
+
6
H
2
→
4
P
H
3
Now notice that they give the amount of hydrogen gas in litres. That will have to be converted to moles. We can use the molar volume of gas,
22.4
L
mol
to do this.
98.2
L
⋅
mol
22.4
L
=
4.38
m
o
l
Now we have to convert grams of phosphorus to moles of phosphorus. Since phosphorus is
P
4
, its molar mass will be
4
(
30.97
)
=
123.88
g
mol
.
11.3
g
⋅
mol
123.88
g
=
0.0912
m
o
l
P
4
Clearly this will run out much before the hydrogen gas, so we can only react this much, therefore, the amount of phosphine produced will be
4
(
0.0912
)
=
0.3648
moles. The final step is to convert this into grams. We can check via the periodic table that the molar mass of phosphine is
34.0
g/
mol
.
0.3648
m
o
l
⋅
34.0
g
mol
=
12.4
g
phosphine