Chemistry, 06.05.2020 08:28, mmsomefood85

Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 0.802 g of methane is mixed with 1.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

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answered: canyonrico05

1.07g

Explanation:

Step 1:

We will begin by writing the balanced equation for the reaction. This is given below:

CH4 + 2O2 —> CO2 + 2H2O

Step 2:

Determination of the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 2 x 32 = 64g

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

Summary:

From the balanced equation above,

16g of CH4 reacted with 64g of O2 to produce 36g of H2O.

Step 3:

Determination of the limiting reactant.

We need to know which of the reactant is limiting the reaction in order to obtain the maximum mass of water.

This is illustrated below:

From the balanced equation above,

16g of CH4 reacted with 64g of O2.

Therefore, 0.802g of CH4 will react with = (0.802 x 64)/16 = 3.21g of O2.

From the above calculations, a higher mass of O2 is needed to react with 0.802g of CH4. Therefore, O2 is the limiting reactant.

Step 4:

Determination of the mass of H2O produced from the reaction.

To obtain the maximum mass of H2O produced, the limiting reactant will be used because it will generate the maximum yield of the product.

From the balanced equation above,

64g of O2 produce 36g of H2O.

Therefore, 1.9g of O2 will produce = (1.9 x 36)/64 = 1.07g of H2O.

The maximum mass of water (H2O) produced by the reaction is 1.07g

answered: Guest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

answered: Guest

the answer is option a.

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Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 0.802 g of methane is mixed with 1.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

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