How many moles are in 75.0g of Na2CO3
1 answer:
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Answer:
0.88 g
Explanation:
Using ideal gas equation to calculate the moles of chlorine gas produced as:-
where,
P = pressure of the gas = 805 Torr
V = Volume of the gas = 235 mL = 0.235 L
T = Temperature of the gas =
R = Gas constant =
n = number of moles of chlorine gas = ?
Putting values in above equation, we get:
According to the reaction:-
1 mole of chlorine gas is produced when 1 mole of manganese dioxide undergoes reaction.
So,
0.01017 mole of chlorine gas is produced when 0.01017 mole of manganese dioxide undergoes reaction.
Moles of = 0.01017 moles
Molar mass of = 86.93685 g/mol
So,
Applying values, we get that:-
<u>0.88 g of should be added to excess HCl (aq) to obtain 235 mL of
at 25 degrees C and 805 Torr.</u>
Answer:
2.24dm³
Explanation:
Given parameters:
Mass of He = 40g
Unknown:
Volume of Helium = ?
Solution:
To solve this problem, we convert the given mass to number of moles.
Number of moles =
molar mass of He = 4g/mol
Number of moles = = 0.1mole
So;
1 mole of gas at rtp occupies a volume of 22.4dm³
0.1 mole of He will occupy a volume of 0.1 x 22.4 = 2.24dm³