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3 years ago

7

Calculate the mass of water formed from the complete combustion of 5.10 g of methane, ch4, in the reaction ch4 (g) + 2 o2 (g) --

> co2 (g) + 2 h2o (g).

1 answer:

Anton [14]3 years ago

4 0

CH₄ + 2O₂ ---> CO₂ + 2H₂O
16g..................................18g×2

16g CH₄ --- 36g H₂O
5,1g CH₄ --- X
X = (5,1×36)/16
X = 11,475g H₂O

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Consider the reaction. 2 Pb ( s ) + O 2 ( g ) ⟶ 2 PbO ( s ) An excess of oxygen reacts with 451.4 g of lead, forming 367.5 g of

Art [367]

Answer : The percent yield of the reaction is, 75.6 %

Solution : Given,

Mass of Pb = 451.4 g

Molar mass of Pb = 207 g/mole

Molar mass of PbO = 223 g/mole

First we have to calculate the moles of Pb.

$\text{ Moles of }Pb=\frac{\text{ Mass of }Pb}{\text{ Molar mass of }Pb}=\frac{451.4g}{207g/mole}=2.18moles$

Now we have to calculate the moles of $PbO$

The balanced chemical reaction is,

$2Pb(s)+O_2(g)\rightarrow 2PbO(s)$

From the reaction, we conclude that

As, 2 mole of $Pb$ react to give 2 mole of $PbO$

So, 2.18 mole of $Pb$ react to give 2.18 mole of $PbO$

Now we have to calculate the mass of $PbO$

$\text{ Mass of }PbO=\text{ Moles of }PbO\times \text{ Molar mass of }PbO$

$\text{ Mass of }PbO=(2.18moles)\times (223g/mole)=486.1g$

Theoretical yield of $PbO$ = 486.1 g

Experimental yield of $PbO$ = 367.5 g

Now we have to calculate the percent yield of the reaction.

$\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }PbO}{\text{ Theoretical yield of }PbO}\times 100$

$\% \text{ yield of the reaction}=\frac{367.5g}{486.1g}\times 100=75.6\%$

Therefore, the percent yield of the reaction is, 75.6 %

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Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

$\Delta T_f=i\times k_f\times m$ ..[1]

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Where:

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$k_f$ = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: $\Delta T_{f,A}$

Molar mass of A = $M_A$

The depression in freezing point of solution with B solute: $\Delta T_{f,B}$

Molar mass of B = $M_B$

$\Delta T_{f,A}>\Delta T_{f,B}$

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

$\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}$

$M_A$

This means compound B has greater molar mass than compound A,

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3 years ago