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.
Answer: 205000000 microliter
Hope this helps!
As,
Water has a pkw=14
so it can be represented as,
[H+] [OH-] = 1*10^-14
If [H+] = 3*10^-5M
[OH-] = (1*10^-14) / ( 3*10^-5)
[OH-] = 3.3*10^-9 M
The balanced chemical reaction describing this decomposition is as follows:
<span>4c3h5n3o9 .............> 6N2 + 12CO2 +10H2O + O2
From the periodic table:
mass of oxygen = 16 grams
mass of nitrogen = 14 grams
mass of hydrogen = 1 gram
mass of carbon = 12 grams
Therefore:
mass of </span><span>C3H5N3O9 = 3(12) + 5(1) + 3(14) + 9(16) = 227 grams
mass of O2 = 2(16) = 32 grams
From the balanced chemical equation:
4(227) = 908 grams of </span>C3H5N3O9 produce 32 grams of O2. Therefore, to know the amount of oxygen produced from 4.5*10^2 grams <span>C3H5N3O9, all we need to do is cross multiplication as follows:
amount of oxygen = (4.5*10^2*32) / (908) = 15.859 grams</span>
