Answer : The reagent present in excess and remains unreacted is, 
Solution : Given,
Moles of 
= 3.00 mole
Moles of = 2.00 mole
Excess reagent : It is defined as the reactants not completely used up in the reaction.
Limiting reagent : It is defined as the reactants completely used up in the reaction.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react with 1 mole of
So, 3.00 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the reagent present in excess and remains unreacted is,
Volume of the tank is 5.5 litres.
Explanation:
mass of the CO2 is given 8.6 grams
Pressure of the gas is 89 Kilopascal which is 0.8762 atm
Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)
R = gas constant 0.0821 liter atmosphere per kelvin)
FROM THE IDEAL GAS LAW
PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)
no of moles = mass/atomic mass
= 8.6/44
= 0.195 moles
now putting the values in equation
V=nRT/P
= 0.195*0.0821*302/ 0.8762
= 5.5 litres.
As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.
this is the formula and answer of this question
Answer:
Na₂CO₃•H₂O
Explanation:
After it is heated, the remaining mass is the mass of sodium carbonate.
30.2 g Na₂CO₃
Mass is conserved, so the difference is the mass of the water:
35.4 g − 30.2 g = 5.2 g H₂O
Convert masses to moles:
30.2 g Na₂CO₃ × (1 mol Na₂CO₃ / 106 g Na₂CO₃) = 0.285 mol Na₂CO₃
5.2 g H₂O × (1 mol H₂O / 18.0 g H₂O) = 0.289 mol H₂O
Normalize by dividing by the smallest:
0.285 / 0.285 = 1.00 mol Na₂CO₃
0.289 / 0.285 = 1.01 mol H₂O
The ratio is approximately 1:1. So the formula of the hydrate is Na₂CO₃•H₂O.
