I’m a paragraph of no less than 125 words describe the steps you can take to effectively organize your writing
1 answer:
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Answer:
7.5 g
Explanation:
There is some info missing. I think this is the original question.
<em>Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.</em>
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Step 1: Write the balanced equation
H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄
Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid
The molar mass of phosphoric acid is 98.00 g/mol.
Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid
The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.
Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate
The molar mass of ammonium phosphate is 149.09 g/mol.
Answer:
0.56L
Explanation:
This question requires the Ideal Gas Law: where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.
Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:
Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means and
Lastly, we must calculate the number of moles of there are. Given 0.80g of
, we will need to convert with the molar mass of
. Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2:
Thus,
Isolating V in the Ideal Gas Law:
...substituting the known values, and simplifying...
So, 0.80g of would occupy 0.56L at STP.