Answer:If the diagram shows 3 m and 6 m the answer would be 2.
Explanation:
Explanation:
The given data is as follows.
50 ml of
, 50 ml of 
And, it is known that at STP 1 mole of a gas occupies 22.4 L. Hence, moles present in 50 ml of gas are as follows.
(As 1 L = 1000 ml)
=
moles
So, according to the given equation
moles of
reacts with
moles of
.
Hence, moles of
is equal to the moles of
and
.
Therefore, moles of
=
moles
1 mole of
= 22.4 L
moles =
= 50 ml of product
Thus, we can conclude that 50 ml of products if pressure and temperature are kept constant.
43.56 grams of are produced if 16g of CH4 reacts with 64g of O2.
Explanation:
Balance equation for the reaction:
CH4 + 2O2⇒ CO2 +2H2O
Data given : mass of CH4 =16 grams atomic mass = 16.04 grams/mole
mass of water 36 gram atomic mass = 18 grams/moles
mass of CO2=? atomic mass = 44.01 grams/mole
number of moles =
equation 1
number of moles in CH4
n = 
= 0.99 moles
Since combustion is done in presence of oxygen hence it is an excess reagent and methane is limiting reagent so production of CO2 depends on it.
From the equation
1 mole of CH4 gave 1 mole of CO2
O.99 moles of CH4 will give x moles of CO2
= 
x = 0.99 moles of carbon dioxide
grams of CO2 = number of moles x atomic mass
= 0.99 x 44.01
= 43.56 grams of CO2 is produced.