The reaction between phosphoric acid and ammonia that produces ammonium phosphate can be written as follows:
3NH3 + H3PO4 ..................> (NH4)3PO4
From the periodic table:
molar mass of nitrogen = 14 grams
molar mass of hydrogen = 1 grams
molar mass of oxygen = 16 grams
molar mass of phosphorus = 30.9 grams
based on this:
molar mass of 3NH3 = 3 (14 + 3(1)) = 51 grams
molar mass of H3PO4 = 3(1) + 30.9 + 4(16) = 97.9 grams
molar mass of (NH4)3PO4 = 3 (14 + 4(1)) + 30.9 + 4(16) = 54 + 30.9 + 64
= 148.9 grams
Therefore, 97.9 grams of phosphoric acid is required to produced 148.9 grams of ammonium phosphate.
Thus, to know the mass of ammonium phosphate produced from 4.9 grams of phosphoric acid, we will simply use cross multiplication as follows:
amount of produced ammonium phosphate = (4.9 x 148.9) / 97.9 = 7.45 g
The answer is 2H2 + O2----> 2H2O
The masses of the components are obtained as;
- Sodium hydrogen carbonate = 3.51 g
- Sodium carbonate = 8.708 g
<h3>What is decomposition?</h3>
The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.
Now putting down the equation of the reaction, we have;
Now, the loss in mass must be due to the carbon dioxide and the water. Hence we obtain the loss in mass to be 10.000 g - 8.708 g = 1.292 g
Mass of sodium hydrogen carbonate = 2 * 88 g/mol * 1.292 g/62 g/mol
= 3.51 g
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Answer:
Accuracy is the closeness to the specific target and precision is the closeness of the measurements to each other.
Answer:
<em>The electrons in an atom can only occupy certain allowed energy levels to a lower one</em>, the excess energy is emitted as a photon of light, with its wavelength dependent on the change in electron energy. This is why an atom can only emit specific wavelengths of light and not every possible wavelength.
