All categories

2 years ago

Please help me i would appreciate it- 50 points and will mark brainliest

Chemistry

1 answer:

4vir4ik [10]2 years ago

5 0

Answer: See below

Explanation:

C3H8 + 502 ----> 3CO2 + 4H2O

1 mol of C₃H₈ reacts to 5 moles of oxygen, in a combustion reaction in order to produce 3 moles of carbon dioxide and 4 moles of water.

First of all, we need to know the limiting reagent:

4.71 g * 1mol / 44g = 0.107 moles of C₃H₈

4.13 g * 1mol/ 32g = 0.129 moles of oxygen

1 mol of C₃H₈ can react to 5 moles of O₂

Then, O moles, must react to (0.107 * 5) /1 = 0.535 moles.

We only have 0.129 moles of O₂ and we need 0.535 moles. We do not have enough oxygen, so it's the limiting reagent.

5 moles of oxygen can produce 3 moles of CO₂

0.129 moles can produce (0.129 * 3) /5 = 0.0774 moles.

Convert the moles to mass : 0.0774 mol * 44g /mol = 3.4056g

Therefore, the maximum mass of CO2 that can be formed is 3.4056g

You might be interested in

An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the

Sauron [17]

Answer:

a) 0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.

b) 0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.

c) Instantaneous rate of the reaction at t=800 s :

Instantaneous rate : $\frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s$

Explanation:

Average rate of the reaction is given as;

$R_{avg}=-\frac{\Delta A}{\Delta t}=\frac{A_2-A_1}{t_2-t_1}$

a.) The average reaction rate between 0.0 s and 1500.0 s:

At 0.0 seconds the concentration was = $A_1=0.184 M$

$t_1=0.0s$

At 1500.0 seconds the concentration was = $A_2=0.016 M$

$t_2=1500 s$

$R_{avg]=-\frac{0.016 M-0.184 M}{1500.0 s-0.0 s}=0.000112 M/s$

0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.

b.) The average reaction rate between 200.0 s and 1200.0 s:

At 0.0 seconds the concentration was = $A_1=0.129 M$

$t_1 =200.0 s$

At 1500.0 seconds the concentration was = $A_2=0.019M$

$t_2=1200 s$

$R_{avg]=-\frac{0.019 M-0.129M}{1200.0s-200.0s}=0.00011 M/s$

0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.

c.) Instantaneous rate of the reaction at t=800 s :

At 800 seconds the concentration was = $A=0.031 M$

$t =800.0 s$

Instantaneous rate : $\frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s$

8 0

4 years ago

In the diagram, which letter represents the energy of the products?

inessss [21]

Answer:

Explanation:

8 0

4 years ago

Read 2 more answers

How many sulfur atoms are present in 25.6 g of Al2(S2O3)3

IgorC [24]

Given the mass of $Al_{2}(S_{2}O_{3})_{3}$ =25.6 g

The molar mass of $Al_{2}(S_{2}O_{3})_{3}$ =390.35g/mol

Converting mass of $Al_{2}(S_{2}O_{3})_{3}$ to moles:

$25.6 g Al_{2}(S_{2}O_{3})_{3}*\frac{1molAl_{2}(S_{2}O_{3}}{390.35 gAl_{2}(S_{2}O_{3}} =0.0656molAl_{2}(S_{2}O_{3}$

Converting mol $Al_{2}(S_{2}O_{3})_{3}$ to mol S:

$0.0656mol Al_{2}(S_{2}O_{3})_{3}*\frac{6molS}{1mol Al_{2}(S_{2}O_{3})_{3}}=0.3936 molS$

Converting mol S to atoms of S using Avogadro's number:

1 mol = $6.022*10^{23}atoms$

$0.3936mol S *\frac{6.022*10^{23}atoms S}{1 mol S}=2.37*10^{23} S atoms$

5 0

3 years ago

Initially, [NH3(g)] = [O2(g)] = 3.60 M; at equilibrium [N2O4(g)] = 0.60 M. Calculate the equilibrium concentration for NH3.

natima [27]

The question is incomplete, here is the complete question:

Consider the reaction $4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)$

Initially, $[NH_3(g)]=[O_2(g)]$ = 3.60 M; at equilibrium $[N_2O_4(g)]=0.60M$ . Calculate the equilibrium concentration for $NH_3$

<u>Answer:</u> The equilibrium concentration of ammonia is 2.8 M

<u>Explanation:</u>

We are given:

Initial concentration of $[NH_3(g)]$ = 3.60 M

Initial concentration of $[O_2(g)]$ = 3.60 M

For the given chemical equation:

$4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)$

Initial: 3.60 3.60

At eqllm: 3.60-4x 3.60-7x 2x 6x

We are given:

Equilibrium concentration of $[N_2O_4(g)]$ = 0.60 M

Evaluating the value of 'x'

$\Rightarrow 2x=0.60\\\\\Rightarrow x=\frac{0.60}{2}=0.2$

So, equilibrium concentration of $NH_3=(3.60-4x)=(3.60-(4\times 0.2))=2.8M$

Hence, the equilibrium concentration of ammonia is 2.8 M

6 0

3 years ago

What will happen to the efficiency of a bike after the gears have been cleaned and the chain has been oiled ?

seropon [69]

Well, figures for efficiency vary a lot, but according to Bicycling Science it’s lubrication that matters most – lubing a dry chain can add 5% to the efficiency.

More interestingly (and I hadn’t read this bit in the book before) it varys a lot depending on gear ratio – bottom gear (22-28) is 99% efficient, top gear (42-11) is 88%. That’s a big difference

7 0

3 years ago