Advertisements
Advertisements
प्रश्न
Ethane burns in oxygen to form CO2 and H2O according to the equation:
`2C_2H_6+7O_2 -> 4CO_2 + 6H_2O`
If 1250 cc of oxygen is burnt with 300 cc of ethane.
Calculate:
1) the volume of `CO_2` formed
2) the volume of unused `O_2`
Advertisements
उत्तर
So, ethane is limiting reagent.
1) 2 x 22400 cc ethane give -> 4 x 22400 cc `CO_2`
∴ 300 cc ethane given → `(4xx22400xx300)/(2xx22400) "cc" Co_2`
= 600 cc CO2
2) For 300 cc Ethane 1050 cc of O2 will be required.
So unused O2 is (1250-1050) = 200 cc
APPEARS IN
संबंधित प्रश्न
Consider the following reaction and based on the reaction answer the questions that follow:
Calculate:
1) the quantity in moles of (NH4)2Cr2O7 if 63gm of(NH4)2Cr2O7 is heated.
2) the quantity in moles of nitrogen formed.
3) the volume in liters or dm3 of N2 evolved at S.T.P.
4) the mass in grams of Cr2O3 formed at the same time
(Atomic masses: H=1, Cr= 52, N=14]
Give the empirical formula of CH3COOH
Give example of compound whose:
Empirical formula is the same as the molecular formula.
Calculate the empirical formula of the compound having 37.6% sodium, 23.1% silicon and 39.3% oxygen.(Answer correct to two decimal places) (O = 16, Na = 23, Si = 28)
The compound A has the following percentage composition by mass: C =26.7%, O = 71.1%, H = 2.2%.
Determine the empirical formula of A.(Answer to one decimal place) (H=1,C=12,O=16)
Determine the empirical formula of the compound whose composition by mass is 42% nitrogen, 48% oxygen and 9% hydrogen.[N=14,O=16,H=1]
The percentage composition of sodium phosphate, as determined by analysis is : 42.1% Na, 18.9% P, 39% of O. Find the empirical formula of the compound.
[H =1, N =14, Na = 23, P = 31, Cl = 35.5, Pt = 195]
A compound of X and Y has the empirical formula XY2. Its vapour density is equal to its empirical formula weight. Determine its molecular formula.
Find the percentage of mass water in Epsom salt MgSO4.7H2O.
A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correctly to 1 decimal place. (H = 1; \[\ce{C}\] = 12; \[\ce{Cl}\] = 35.5)
