Answer:
5.67 g OF WATER WILL BE FORMED WHEN 13.7 g OF MnO2 REACTS WITH HCl GAS.
Explanation:
EQUATION FOR THE REACTION
Mn02 + 4HCl --------> MnCl2 + Cl2 + 2H2O
From the balanced reaction between manganese oxide and hydrogen chloride gas;
1 mole of MnO2 reacts to form 2 mole of water
At STP, the molecular mass of the sample is equal to the mole of the substance. So therefore:
(55 + 16 * 2) g of MnO2 reacts to form 2 * ( 1 *2 + 16) g of water
(55 + 32) g of MnO2 reacts to form 2 * 18 g of water
87 g of MnO2 reacts to form 36 g of water
If 13.7 g of MnO2 were to be used?
87 g of MnO2 = 36 g of H2O
13.7 g of MnO2 = ( 13.7 * 36 / 87) g of water
= 493.2 / 87 g of water
Mass of water = 5.669 g of water
Approximately 5.67 g of water will be formed when 13.7 g of manganese oxide reacts with excess hydrogen chloride gas.
<span>Given in the question-
1 mole of cyclohexanol = > 1 mole of cyclohexene
Molar mass 100.16 g/mol
moles of cyclohexanol = .240 / 100.16= 0.002396 moles
Molar mass 82.143 g/mol
moles of cyclohexene formed @100 % yield = 0.002396
Molar mass 82.143 g/mol
mass of cyclohexene @ 100 % = .002396 x 82.143 = 0.197g
bur we have .138g
so % yield = .138 / .197 = 70.0 %
Ans- 70 percentage yield of cyclohexene.</span>
