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How much Mercury (II) Oxide is needed to produce 198g of Oxygen?

Sagot :

Answer:

The Same Of this Equetion Answer

Explanation:

First we need to convert the chemical equation that you wrote out in words into a chemical equation with compounds and coefficients. The balanced chemical equation looks like this:

2 HgO (s) --> 2 Hg (l) + O2 (g)

Now we need to convert the 55.7 g of HgO into moles of HgO. To do this, we find the molar mass of HgO. The periodic table tells us that the atomic mass of one Hg is 200.59. The periodic table also tells us that the atomic mass of one O is 15.999. Thus, the molar mass of HgO is 200.59 + 15.999 = 216.589 grams. That is, one mole of HgO = 216.589 grams of HgO.

We need to figure out how many moles are in 55.7 grams of HgO. We calculate this by doing 55.7 divided by 216.589. The calculator says that 55.7 / 216.589 = 0.25716911. So 55.7 grams of HgO is equal to 0.25716911 mole of HgO.

Now we use the coefficients in the balanced equation to convert from moles of HgO to moles of O2. The equation to convert looks like this:

moles of HgO x coefficient of O2

coefficient of HgO

In case the formatting messes up, this is moles of HgO multiplied by (coefficient of O2 / coefficient of HgO). We already know the moles of HgO (it's the 0.25716911 moles of HgO that we just calculated). The coefficients come from the balanced equation; here it is again: 2 HgO (s) --> 2 Hg (l) + O2 (g) . The coefficient of HgO is 2, and the coefficient of O2 is 1. (Remember that if there is no visible coefficient, the coefficient is 1.) If we put these numbers in the equation to convert, the equation to convert looks like this:

0.25716911 moles of HgO x 1

2

If you can't see this, we do 0.25716911 x (1/2). The calculator says that 0.25716911 x (1/2) = 0.12858455. Thus, we have 0.12858455 moles of O2. Remember that the question asks for the volume of O2. Now we need to use the ideal gas equation to find the volume of the 0.12858455 moles of O2. The ideal gas equation is:

PV = nRT

Remember that P is pressure in atm, V is the volume of the gas in liters, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin. So far we know that P = 1 atm (the problem gave us this information), V is what we want to find, n = 0.12858455 moles of O2, and T = 25 degrees Celsius. But we must have the temperature in Kelvin, so we convert from Celsius to Kelvin with the equation

K = C + 273.15. Since the temperature is 25 degrees Celsius, K = 25 + 273.15 = 298.15. So the temperature is 298.15 K. Finally, R = 0.082057 L*atm/(K*mol). (You can look up the value for R in a table or chart; you don't need to memorize it.) Once we put these values into the ideal gas equation, the ideal gas equation looks like this:

1*V = (0.12858455)*(0.082057)*(298.15)

Since 1*V is equal to V, we can just write V instead of 1*V. So our equation looks like this:

V = (0.12858455)*(0.082057)*(298.15)

Now we just multiply the numbers together. The calculator says that (0.12858455)*(0.082057)*(298.15) = 3.14589. Remember that the correct unit for volume is liters. Thus, the volume of O2 is 3.14589 L. We should round to 3 significant figures because the problem gave us the number 55.7 (which has 3 significant figures). If we round 3.14589 to 3 significant figures, we get 3.15. Thus, the answer is 3.15 L of O2.