Answer:
Question Clear
Explanation:
Make sure your question is clear.
Answer:
Sodium oxide is the product
Explanation:
4Na+O2->2Na2O
<span>0.925 grams if using hydrochloric acid in the reaction.
0.462 grams if using sulfuric acid in the reaction.
0.000 grams if using nitric acid in the reaction.
Assuming you're using HCl or a similar acid for this reaction, the equation for the reaction is:
Zn + 2 HCl ==> ZnCl2 + H2
So each mole of zinc used, produces 1 mole of hydrogen gas, or 2 moles of hydrogen atoms. So we need to look up the atomic weights of both zinc and hydrogen.
Atomic weight zinc = 65.38
Atomic weight hydrogen = 1.00794
Moles zinc = 30.0 g / 65.38 g/mol = 0.458855919 mol
Since we produce 2 moles of hydrogen atoms per mole of zinc, multiply by 2 and the atomic weight of hydrogen to get the mass of hydrogen produced. So
0.458855919 * 2 * 1.00794 = 0.92499847 grams.
Rounding to 3 significant figures gives 0.925 grams.
To show the assumption of the acid used, the balanced equation for sulfuric acid would be
Zn2 + H2SO4 ==> Zn(SO4)2 + H2
Which means that for every mole of zinc used, 1 mole of hydrogen gas is generated (half that produced via hydrochloric acid).
If nitric acid were used, the reaction is
4Zn + 10HNO3 ==> 4Zn(NO3)2 + N2O + 5H2O
Which means that NO hydrogen gas is generated.
The only justification for assuming hydrochloric acid is used is that it's a fairly common acid that's easy to obtain. But as shown above with 2 alternative acids, the amount of hydrogen gas generated is very dependent upon the exact chemical reaction occurring and asking "How many grams of hydrogen are produced if 30.0 g of zinc reacts?" is a rather silly question unless you specify EXACTLY what the reaction is.</span>
Answer:
The molar mass is determined by applying the Ideal Gas Law, PV = nRT, where P is the pressure (in atm), V is the volume (in L), n is the number of moles of gas, R is the universal gas constant (0.08206 L∙atm/mol∙K), and T is the temperature (in K).
Hope this helps! :)
You determine the correct number of significant figure of a measurement in a graduated cylinder by looking at the smallest division on the graduated cylinder. If the division is divided up to the ones place, you can still read the half of that division or the .50 measurement. So, you can have until the tenths place in your measurement.
