What is the ratio by mass of oxygen to carbon when 32 g of oxygen combine with 12 g of carbon?

sleet_krkn [62]

Answer:

Magnesium and calcium belong to the second group i. e. alkaline earth metals. They are known as earth metals because they are extracted from the earth. They are very reactive elements. Their reactivity increases when we go from top to bottom because the outermost electrons goes farther from the nucleus i. e. atomic radius increases so less energy is needed for its removal.

8 0

3 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.

Moles of $H_2O$ reacted = 1.05 mol $Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}$

= 6.31 mol $H_2O$

The mass of $H_2O$ is,

$m_{H_{2} O}$ = 6.31 mol $\times$ 18.02g/ mol

= 114g

On subtracting, the mass of $H_2O$ reacted from the given mass of $H_2O$ is,

$m_{H_2O}$ = (131-114)g

= 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

#SPJ4

8 0

2 years ago

What do radio waves and microwaves have in common?

Anastasy [175]

1) A
2)A
3)C
4)A
That should be all of them. Hope this helps?

6 0

3 years ago

Consider the reaction: 2 H2O (g)-->2 H2 (g) + O2 (g). ΔH=483.6 Kj/mol. If 2 moles of H2O (g) are converted H2(g) and O2(g) ag

denis-greek [22]

DE = dH - PdV

2 H2O(g) → 2 H2(g) + O2(g) 

You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. 

1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. 

Since it is done under constant temp and pressure that means the volume change will be equal to the volume of 1 mole of gas 

2 moles reacts to form 3 moles 

The gas equation is 

PV = nRT 
P = pressure 
V = volume (unknown) 
n = moles (1) 
R = gas constant = 8.314 J K^-1 mol^-1 
- the gas constant is different for different units of temp and pressure (see wikki link) in this case temp and pressure are constant, and we want to put the result in an equation that has Joules in it, so we select 8.314 JK^-1mol^-1) 
T = temp in Kelvin (kelvin = deg C + 273.15 
So T = 403.15 K 

Now, you can see that PV is on one side of the equation, and we are looking to put PdV in our dE equation. So we can say 

dE = dH -dnRT (because PV = nRT) 

Also, since the gas constant is in the unit of Joules, we need to convert dH to Joules 

dH = 483.6 kJ/mol = 483600 Joules/mol 

dE = 483600 J/mol - (1.0 mol x 8.314 J mol^-1K-1 x 403.15 K) 
dE = 483600 J/mol - 3351.77 J 
dE = 480248.23 J/mol 
dE = 480.2 kJ/mol

5 0

3 years ago

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Im not doing the math-

BigorU [14]

The average speed of the blocks are 0.36 m/s.
Explanation:
Average speed is defined as the ratio of distance covered per unit time. So if it is said that blocks are pulled to 0.9 m in the right side. This means the blocks cover a distance of 0.9 m from the origin and that distance is covered in 2.5 s. Thus, the average speed can be calculated from the change in speed with respect to time. As at time t = 0 , the speed is also zero, and at time t = 2.5 s , the speed will be
Since, in this case, the speed is equal to the average speed of blocks. So the average speed of the blocks will be 0.36 m/s.

3 0

3 years ago

Read 2 more answers