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2 years ago

Need help with this please thanks

Chemistry

1 answer:

kolbaska11 [484]2 years ago

5 0

Answer: 1. $2H_2+O_2\rightarrow 2H_2O$

2. $P_4+3O_2\rightarrow 2P_2O_3$

3. $N_2+3H_2\rightarrow 2NH_3$

4. $2K+Cl_2\rightarrow 2KCl$

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The given equations are balanced as:

1. $2H_2+O_2\rightarrow 2H_2O$

2. $P_4+3O_2\rightarrow 2P_2O_3$

3. $N_2+3H_2\rightarrow 2NH_3$

4. $2K+Cl_2\rightarrow 2KCl$

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2 years ago

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3 years ago

A 3.00-L flask is filled with gaseous ammonia, NH3. The gas pressure measured at 27.0 ∘C is 2.55 atm . Assuming ideal gas behavi

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Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.

Solution :

Using ideal gas equation,

$PV=nRT\\\\PV=\frac{w}{M}\times RT$

where,

n = number of moles of gas

w = mass of ammonia gas = ?

P = pressure of the ammonia gas = 2.55 atm

T = temperature of the ammonia gas = $27^oC=273+27=300K$

M = molar mass of ammonia gas = 17 g/mole

R = gas constant = 0.0821 L.atm/mole.K

V = volume of ammonia gas = 3.00 L

Now put all the given values in the above equation, we get the mass of ammonia gas.

$(2.55atm)\times (3.00L)=\frac{w}{17g/mole}\times (0.0821L.atm/mole.K)\times (300K)$

$w=5.28g$

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3 years ago

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3 years ago

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

shepuryov [24]

The reaction forms 98.76 g AlCl_3.

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

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

Step 1. Gather all the information 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

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Step 2. Calculate the moles of each reactant

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 limiting reactant

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

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

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

Cl_2 is the limiting reactant because it gives the smaller amount of AlCl_3.

Step 4. Calculate the mass of AlCl_3.

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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2 years ago