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3 years ago

How many grams of CO are needed to react with an excess of Fe2O3 to produce (210.3 g Fe) 209.7 g Fe? Show your work.

2 answers:

6 0

The balanced chemical reaction is given as:

<span>3CO (g) + Fe2O3 (s) --> 2Fe (s) + 3CO2 (g)

We are given the amount of the product to be produced from the reaction. We use this as the starting value for the calculations. We do as follows:

209.7 g Fe ( 1 mol / 210.3 g Fe ) ( 3 mol CO / 2 mol Fe ) (28.01 g / 1 mol) = 41.90 g CO needed</span>

Fed [463]3 years ago

4 0

Explanation:

$3CO(g)+Fe_2O_3(s)\rightarrow 2Fe(s)+3CO_2(g)$

1)Mass of CO when 210.3 g of Fe produced.

Number of moles of $Fe$ in 210.3 g=

$\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}$

$=\frac{210.3}{55.84 g/mol}=3.76 mol$

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.76 moles of Fe will be obtained from : $\frac{3}{2}\times 3.76 moles$ of CO that is 5.64 moles.

Mass of CO in 5.64 moles =

$\text{Number of moles}\times \text{molar mass of CO}=5.46\times 28 g/mol=157.92 g$

2)Mass of CO when 209.7 g of Fe produced.

Number of moles of $Fe$ in 209.7 g=

$\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}$

$=\frac{209.7}{55.84 g/mol}=3.75 mol$

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.75 moles of Fe will be obtained from : $\frac{3}{2}\times 3.75 moles$ of CO that is 5.625 moles.

Mass of CO in 5.625 moles =

$\text{Number of moles}\times \text{molar mass of CO}=5.625\times 28 g/mol=157.5 g$

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3 years ago

A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r

Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

$P\cdot V = n\cdot R\cdot T$ ,

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of gas particles in the gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume $V$ of the gas. Rearrange the ideal gas equation for volume:

$V = \dfrac{n \cdot R \cdot T}{P}$ .

Both the temperature of the gas, $T$ , and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

$T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K}$ ,
$T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}$ .

The volume of the gas is proportional to its temperature if both $n$ and $P$ stay constant.

$n$ won't change unless the balloon leaks, and
consider $P$ to be constant, for calculations that include $T$ .

$V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}$ .

Now, keep the temperature at $T_1 =258.65\;\text{K}$ and change the pressure on the gas:

$P_1 = 0.995\;\text{atm}$ ,
$P_2 = 0.720\;\text{atm}$ .

The volume of the gas is proportional to the reciprocal of its absolute temperature $\dfrac{1}{T}$ if both $n$ and $T$ stays constant. In other words,

$V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}$

(3 sig. fig. as in the question.).

See if you get the same result if you hold $T$ constant, change $P$ , and then move on to change $T$ .

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