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vlada-n [284]
3 years ago
11

20.00 g of aluminum (Al) reacts with 78.78 grams of molecular chlorine (Cl2), all of each reaction is completely consumed and as

ingle product is formed. What is the mass (in grams) of the product?
Chemistry
1 answer:
shepuryov [24]3 years ago
4 0

The reaction forms 98.76 g AlCl_3.  

We have the masses of two reactants, so this is a <em>limiting reactant problem</em>.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Step 1. Gather all the information</em> in one place with molar masses above the formulas and everything else below them.  

M_r: ___26.98 _70.91 __133.34

________2Al + 3Cl_2 → 2AlCl_3

Mass/g: 20.00 _78.78

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>  

Moles of Al = 20.00 g Al × (1 mol Al /26.98 g Al) = 0.741 29 mol Al

Moles of Cl_2 = 78.78 g Cl_2 × (1 mol Cl_2 /70.91 g Cl_2) = 1.11 10 mol Cl_2

Step 3. Identify the <em>limiting reactant</em>  

Calculate the moles of AlCl_3 we can obtain from each reactant.  

<em>From Al</em>: Moles of AlCl_3 = 0.741 29 mol Al × (2 mol AlCl_3/2 mol Al) = 0.741 29 mol AlCl_3

<em>From Cl_2</em>: Moles of AlCl_3 = 1.11 10 mol Cl_2 × (2 mol AlCl_3/3 mol Cl_2) = 0.740 66 mol AlCl_3

<em>Cl_2 is the limiting reactant</em> because it gives the smaller amount of AlCl_3.

<em>Step 4</em>. Calculate the <em>mass of AlCl_3</em>.

Mass = 0.740 66 mol AlCl_3 × 133.34 g/1 mol AlCl_3 = 98.76 g AlCl_3

The reaction produces 98.76 g AlCl_3.

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En un experimento hacemos reaccionar 12 g de carbono con 32 g de oxígeno para formar dióxido de carbono. Razona si podemos saber
max2010maxim [7]

Answer:

La masa de óxido de carbono iv formado es 44 g.

Explanation:

En esta pregunta, se nos pide calcular la masa de óxido de carbono iv formado a partir de la reacción de masas dadas de carbono y oxígeno.

En primer lugar, necesitamos escribir una ecuación química equilibrada.

C + O2 → CO2

De la ecuación, 1 mol de carbono reaccionó con 1 mol de oxígeno para dar 1 mol de óxido de carbono iv.

Ahora, si marca las masas en la pregunta, verá que corresponde a la masa atómica y la masa molar de la molécula de carbono y oxígeno, respectivamente. ¿Qué indica esto?

Como tenemos una relación molar de 1: 1 en todo momento, lo que esto significa es que la masa de óxido de carbono iv producida también es la misma que la masa molar de óxido de carbono iv.

Por lo tanto, procedemos a calcular la masa molar de óxido de carbono iv Esto es igual a 12 + 2 (16) = 12 + 32 = 44 g Por lo tanto, la masa de óxido de carbono iv formado es 44 g

5 0
3 years ago
Assume a gasoline is isooctane, which has a density of 0.692 g/ml. What is the mass of 3.8 gal of the gasoline (1 gal = 3.78 l)?
sveticcg [70]

Density is the ratio of mass to the volume.

The mathematical expression is given as:

density=\frac{mass}{volume}

Now, density of isooctane = 0.692 g/mL

Volume  = 3.8 gal

Since, 1 gallon = 3.78 L

So, 3.8 gal = 3.78 L\times 3.8

= 14.364 L

As, 1 L = 1000 mL

Therefore, 14.364 L= 14.364 L\times 1000 mL

Volume in mL = 14364 mL

Put the values,

0.692 g/mL=\frac{mass}{14364 mL}

m = 0.692 g/mL\times 14364 mL

= 9939.888 g

Hence, mass of 3.8 gal of the gasoline is 9939.888 g.



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Blababa [14]

Answer:

b

Explanation:

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Rasek [7]
An ideal gas differs from a real gas in that the molecules of an ideal gas have no attraction for one another. 
An ideal gas is defined as one in which collisions between atoms or molecules are perfectly elastic and in which there are no inter-molecular attractive forces. A real gas on the other hand is a gas that does not behave as an ideal gas due to interactions between gas molecules. Particles in a real gas have a real volume since real gases are made up of molecules or atoms that typically take up some space even though they are extremely small. 
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Answer:

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