Convert 125m^3 to km^3

Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc

max2010maxim [7]

Answer: The percent yield of the water is 31.98 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For methane:

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol$

For oxygen gas:

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol$

The chemical equation for the combustion of methane is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = $\frac{1}{2}\times 0.45=0.225mol$ of methane

As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = $\frac{2}{2}\times 0.45=0.45$ moles of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

$0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%$

Hence, the percent yield of the water is 31.98 %

4 0

3 years ago

3. What methods are you using to test this hypothesis? Outline the steps of the procedure in full sentences.

Ulleksa [173]

Answer:

AHYPOTHISES IS A GUESS

Explanation:

4 0

3 years ago

WILL GIVE BRAINLIEST!! Determine the volume of a container that holds 2.4 mol of gas at STP.

kykrilka [37]

Answer:

2.4 mol x 22.4 liter = 53.76 liters

1 mole

Explanation:

5 0

3 years ago

The net ionic equation for the reaction between aqueous sulfuric acid and aqueous sodium hydroxide is ________.

Dennis_Churaev [7]

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

2H⁺ + SO₄²⁻ + 2Na⁺ + 2OH⁻ → 2Na⁺ + SO₄²⁻ + 2H₂O

H⁺ + OH⁻ → H₂O (the net ionic equation)

5 0

4 years ago

Calculate the fractional saturation for hemoglobin when the partial pressure of oxygen is 58 mm Hg. Assume hemoglobin is 50% sat

svetlana [45]

Answer:

0.8988

Explanation:

To calculate the fractional saturation of hemoglobin , the formula used is

YO_2= $\frac{(pO_2)^n}{(p50)^n+(pO_2)^n}$

now putting the values

YO_2= $\frac{58^3}{28^3+58^3}$

= $\frac{195112}{21952+195112}$

=0.8988

Therefore, fractional saturation of hemoglobin= 0.8988

where

YO_2= fractional saturation of hemoglobin

pO_2= partial pressure of oxygen

p50= is the pO_2 at which hemoglobin is 50% saturated

4 0

3 years ago