All categories

2 years ago

4 b + 3 o2 → 2 b2o3 if 8 moles of b and 4 moles of o2 are allowed to react, how many moles of b2o3 can be formed?

Chemistry

1 answer:

raketka [301]2 years ago

6 0

The number of moles of b2o3 that will be formed is determined as 4 moles.

<h3>Limiting reagent</h3>

The limiting reagent is the reactant that will be completely used up.

4 b + 3O₂ → 2b₂O₃

from the equation above;

4 b ------------> 2 b₂O₃

2b ------------> b₂O₃

2 : 1

3O₂ -------------> 2b₂O₃

3 : 2

b is the limiting reagent, thus, the amount of b2o3 to be formed is calculated as;

4 b ------------> 2 moles of b2o3

8 moles -------> ?

= (8 x 2)/4

= 4 moles

Thus, the number of moles of b2o3 that will be formed is determined as 4 moles.

Learn more about limiting reactants here: brainly.com/question/14222359

#SPJ1

You might be interested in

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

Bond [772]

Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

3 0

3 years ago

In naming the compound PCl5, the prefix used with the second element is ____________________.

Oksanka [162]

Answer:

$\boxed {\boxed {\sf Hepta}}$

Explanation:

We are given the formula:

$PCl_5$

This is a molecular formula, because it contains nonmetals.

1. Name the first element

The first element is phosphorous (P). Since this is the first element and there is only one, we don't need a prefix.

Phosphorous

2. Second element

The second element is chlorine (Cl). It has a subscript of 5, so we must add the prefix of <u>hepta</u>-.

Phosphorous heptachlorine

Add the ending of -ide.

Phosphorous heptachloride

The prefix used for the second element is hepta. The compound name is phosphorous heptachloride.

6 0

3 years ago

A chlorine atom has a diameter of 175 picometers (pm). Please write the diameter, in metered, in scientific notation.

Flauer [41]

Picometre is equal to 1·10⁻¹²m
175·10⁻¹² m = 1,75·10⁻¹⁰ m

7 0

3 years ago

Imagine that you have a 5.00 L gas tank and a 3.50 L gas tank. You need to fill one tank with oxygen and the other with acetylen

Lorico [155]

Answer:

77.14 atm of pressure should be of an acetylene in the tank.

Explanation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

According to reaction, 2 moles of acetylene reacts with 5 moles of oxygen.

Moles of oxygen= $n_1$

Moles of acetylene = $n_2$

$\frac{n_1}{n_2}=\frac{5 mol}{2 mol}=\frac{5}{2}$

Volume of large tank with oxygen gas, $V_1 = 5.00 L$

Pressure of the oxygen gas inside the tank = $P_1=135 atm$

$RT=\frac{P_1V_1}{n_1}$ ..[1]

Volume of small tank with acetylene gas , $V_2= 3.50 L$

Pressure of the acetylene gas inside the tank = $P_2=?$

$RT=\frac{P_2V_1}{n_2}$ ..[2]

Considering both the gases having same temperature T, [1]=[2]

$\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}$

$P_2=\frac{P_1V_1\times n_2}{V_2\times n_1}$

$=\frac{135 atm\times 5.00 L\times 2}{3.50 L\times 5}=77.14 atm$

77.14 atm of pressure should be of an acetylene in the tank.

8 0

3 years ago

Explain how carbon’s bonding ability makes it unique.

STatiana [176]

$\huge\fcolorbox{red}{pink}{Answer ♥}$

The carbon atom is unique among elements in its tendency to form extensive networks of covalent bonds not only with other elements but also with itself. ... Moreover, of all the elements in the second row, carbon has the maximum number of outer shell electrons (four) capable of forming covalent bonds.

Hope it helps uh ✌️✌️✌️

Gud mrng

7 0

3 years ago