1.Find the molarity of 4g NAOH in 250ml of solution?
2 .50g N2 react with 10g of H2 to from ammonia according to the equation N2+3H2➡️2NH3. Identify the limiting reagents and ammonia formed
3.A welding fuel has contains carbon and hydrogen only burning app small sample of oxygen gives 3.38 gram carbon dioxide 0.6 90 gram of water no other products a volume of 10.0 litre (measured at STP close )of this welding gas found two pound to wiegh 11.6 gram calculate
1 empirical formula
2. Molar mass of the gas and
3 molecular formula
4 .Calcium carbonate reacts with aqueous HCL to give cacl2 and CO2 according to the reaction carbonate + 2 HCL gives cacl2 + CO2 + H2O. What mass of caco3 is required to react completely with 25 ml of 0. 75 m HCL?
5.
Chlorine is prepared in the laboratory by treating Magnus dioxide with aqueous hydrochloric acid according to the reaction 4 HCL+ mno2 gives 2 H2O+ mncl2+ cl2 . How many grams of HCL react with 5.0 gram of Magnus dioxide
Answers
Answer:
1) The molarity of this solution is 0.4.
2) N2+3H2=2NH3
28gm. 6gm. 34gm
28gm N2 require 6g.m
50g.m will require (6/28)×50=10.7g.m H2
but we have only 10gm of H2
so H2 is limiting reagent in rxn
H2 will decide the formation of NH3
6gm of H2 produce=34gm
10g.m will produce=(34/6)×10=56.6gm
NH3
_ _ _ _ _ _ _ _ _ _ _ _ _ _
limiting reagent is H2
Amount of NH3 produce=56.6gm
3) Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is
12/44 ×3.38=0.92g
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is
2/18 ×0.690=0.077g
The percentage of C is 0.92/0.92+0.077 ×100=92.3%
The percentage of H is 0.077/0.92+0.077 ×100=7.7%
(i) The number of moles of carbon=92.2/12=7.7
The mole ratio C:H=7.7/7.7=1:1
Hence, the empirical formula of the welding fuel gas is CH.
(ii) The weight of 10.0 L of gas at N.T.P is 11.6 g.
22.4 L pf gas at N.T. P will weigh 11.6/10.0 ×22.4=26g/mol
This is the molar mass.
(iii) Empirical formula mass is 12+1=13.
Molecular mass is 26.
The ratio, n of the molecular mass to empirical formula mass is 26/13=2
The molecular formula is 2(CH)=C2H2.
4) The mass of HCl is 0.75 × 36.5/1000 ×25=0.684g.
1 mole of calcium carbonate reacts with 2 moles of HCl.
Hence, the mass of calcium carbonate that will react completely with 0.684 g of HCl is
100/2×36.5 ×0.684=0.936g.
5)According to the given equation:
MnO2 +4HCl→MnC2+Cl2+2H2O
1 mole of MMnO reacts with 4 mole of HCl or 87g MnO2 reacts with 146g HCl
∴ 5g MnO2 Will react with 146/87 ×5g HCl = 8.39 HCl.
Explanation:
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