Chemistry, 27.06.2019 19:20 robert7248

Given the following equation: 2 c4h10 13 o2 > 8 co2 10 h20 + how many grams of co2 are produced if 12.4 grams of c4h10 reacts with 56.9 grams of o2?

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Answer from: jbearallen9557

Answer : The mass of CO_2 produced will be, 37.488 grams.

Explanation : Given,

Mass of $C_4H_{10}$ = 12.4 g

Mass of O_2 = 56.9 g

Molar mass of $C_4H_{10}$ = 58 g/mole

Molar mass of O_2 = 32 g/mole

Molar mass of CO_2 = 44 g/mole

First we have to calculate the moles of $C_4H_{10}$ and O_2 .

$\text{Moles of }C_4H_{10}=\frac{\text{Mass of }C_4H_{10}}{\text{Molar mass of }C_4H_{10}}=\frac{12.4g}{58g/mole}=0.213moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{56.9g}{32g/mole}=1.778moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_4H_{10}$ react with 13 mole of O_2

So, 0.213 moles of $C_4H_{10}$ react with $\frac{13}{2}\times 0.213=1.385$ moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and $C_4H_{10}$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of CO_2 .

As, 2 moles of $C_4H_{10}$ react to give 8 moles of CO_2

So, 0.213 moles of $C_4H_{10}$ react to give $\frac{8}{2}\times 0.213=0.852$ moles of CO_2

Now we have to calculate the mass of CO_2 .

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=(0.852mole)\times (44g/mole)=37.488g$

Therefore, the mass of CO_2 produced will be, 37.488 grams.

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

idk i’m struggling too