Answer:
27%
Explanation:
Hello,
The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

Finally, the percent yield turns out into:

%
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Iron(II) oxide also refers to a family of related non-stoichiometric compounds, which are typically iron deficient with compositions ranging from Fe0.84O to Fe0.95O.
Answer:

Explanation:
The pressure is constant, so we can use Charles' Law.

Data:
V₁ = 1.92 × 10³ L; T₁ = 20 °C
V₂ = ?; T₂ = 68 °C
Calculations:
(a) Convert temperatures to kelvins
T₁ = (20 + 273.15) K = 293.15 K
T₂ = (68 + 273.15) K = 341.15 K
(b) Calculate the volume

The new volume of the gas is
.
Answer: First, here is the balanced reaction: 2C4H10 + 13O2 ===> 8CO2 + 10H2O.
This says for every mole of butane burned 4 moles of CO2 are produced, in other words a 2:1 ratio.
Next, let's determine how many moles of butane are burned. This is obtained by
5.50 g / 58.1 g/mole = 0.0947 moles butane. As CO2 is produced in a 2:1 ratio, the # moles of CO2 produced is 2 x 0.0947 = 0.1894 moles CO2.
Now we need to figure out the volume. This depends on the temperature and pressure of the CO2 which is not given, so we will assume standard conditions: 273 K and 1 atmosphere.
We now use the ideal gas law PV = nRT, or V =nRT/P, where n is the # of moles of CO2, T the absolute temperature, R the gas constant (0.082 L-atm/mole degree), and P the pressure in atmospheres ( 1 atm).
V = 0.1894 x 0.082 x 273.0 / 1 = 4.24 Liters.
Explanation: