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anastassius [24]
3 years ago
12

If you are given a chemical equation and specific amounts for each reactant in grams, how would you determine how much product c

an possibly be made? If you are given a chemical equation and specific amounts for each reactant in grams, how would you determine how much product can possibly be made? 1. Determine which reactant is the limiting reactant. 2. Using conversion factor, convert grams of the limiting reactant to grams of the product. 1. Determine which reactant is the limiting reactant. 2. Convert each reactant into moles of the product. 3. Convert the moles of product, from the limiting reactant, to grams. 1. Convert each reactant into moles of the product. 2. Determine which reactant is the limiting reactant. 3. Convert the moles of product, from the limiting reactant, to grams.
Chemistry
1 answer:
V125BC [204]3 years ago
8 0

Answer:

<em>If you are given a chemical equation and specific amounts for each reactant in grams, you have to follow these steps, in order, to determine how much product can possilby be made:</em>

  • <u><em>1. Convert each reactant into moles of the product. </em></u>
  • <u><em>2. Determine which reactant is the limiting reactant. </em></u>
  • <u><em>3. Convert the moles of product, from the limiting reactant, to grams</em></u><u>.</u>

Justification:

The balanced chemical equation, through the coefficients, represents the proportions, in terms of mole ratios, in that each reactant chemically react with each other to form each product.

So, if you are given specific aomunts for each reactant in grams, you must start by converting each reactant amount into mole numbers, using the molar mass of each one.

Then, using the theoretical mole ratios (from the coefficients of the balanced chemical equation) you can determine how much of each product can be produced, in terms of moles, first starting with the complete consumption of one of the reactants, and then doing the same with each of the other reactants, one by one.

The reactant that yields the least amount of product is the limiting reactant, since it will be consumed completely once that amount of product is obtainded, while the other reactants will be in excess.

Now, that you have the number of moles that can be produced from the limiting reactant, you can convert it into grams of product, just using the molar mass of the same.

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What is the term for an irregular way a mineral breaks apart?
kari74 [83]

Answer:

That would be fracture

Explanation:

A way that a mineral breaks apart curved at  the edges or split into uneven pieces

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3 years ago
How many grams of KCI can be dissolved in 63.5. g of water at 80
Masteriza [31]

Answer:

35.8 g

Explanation:

Step 1: Given data

Mass of water: 63.5 g

Step 2: Calculate how many grams of KCl can be dissolved in 63.5. g of water at 80 °C

Solubility is the maximum amount of solute that can be dissolved in 100 g of solute at a specified temperature. The solubility of KCl at 80 °C is 56.3 g%g, that is, we can dissolve up to 56.3 g of KCl in 100 g of water.

63.5 g Water × 56.3 g KCl/100 g Water = 35.8 g KCl

8 0
3 years ago
The extraction of pure copper is expensive. Give one reason why
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3 years ago
Read 2 more answers
The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 °c is 4.48. what is the value of ka for hbro?
Alex Ar [27]

____________________________________________________

Answer:

Your answer would be a). 2.0 × 10-9

____________________________________________________

Work:

In your question the "ph" of a 0.55 m aqueous solution of hypobromous acid temperature is at 25 degrees C, and it's "ph" is 4.48.

You would use the ph (4.48) to find the ka for "hbro"

[H+]

=

10^-4.48

=

3.31 x 10^-5 M

=

[BrO-]

or: [H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-]

Then you would find ka:

(3.31 x 10^-5)^2/0.55 =2 x 10^-9

____________________________________________________

<em>-Julie</em>

6 0
3 years ago
An air tight freezer measures 4 mx 5 m x 2.5 m high. With the door open, it fills with 22 °C air at 1 atm pressure.
____ [38]

Answer:

(a) Density of the air = 1.204 kg/m3

(b) Pressure = 93772 Pa or 0.703 mmHg

(c) Force needed to open the door =  15106 N

Explanation:

(a) The density of the air at 22 deg C and 1 atm can be calculated using the Ideal Gas Law:

\rho_{air}=\frac{P}{R*T}

First, we change the units of P to Pa and T to deg K:

P=1 atm * \frac{101,325Pa}{1atm}=101,325 Pa\\\\ T=22+273.15=293.15^{\circ}K

Then we have

\rho_{air}=\frac{P}{R*T}=\frac{101325Pa}{287.05 J/(kg*K)*293.15K} =1.204 \frac{kg}{m3}

(b) To calculate the change in pressure, we use again the Ideal Gas law, expressed in another way:

PV=nRT\\P/T=nR/V=constant\Rightarrow P_{1}/T_{1}=P_{2}/T_{2}\\\\P_{2}=P_{1}*\frac{T_{2}}{T_{1}}=101325Pa*\frac{7+273.15}{22+273.15}=101,325Pa*0.9254=93,772Pa\\\\P2=93,772 Pa*\frac{1mmHg}{133,322Pa}= 0.703 mmHg

(c) To calculate the force needed to open we have to multiply the difference of pressure between the inside of the freezer and the outside and the surface of the door. We also take into account that Pa = N/m2.

F=S_{door}*\Delta P=2m^{2} *(101325Pa - 93772Pa)=2m^{2} *7553N/m2=15106N

7 0
3 years ago
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