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4 years ago

5. 1.00 mol HNO3 is treated with 4.47 g of magnesium. Calculate the number of moles of

Chemistry

1 answer:

ruslelena [56]4 years ago

3 0

Answer:

The balanced equation is:

2 HNO3 + Mg ---> Mg(NO3)2 + H2

From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg

Moles of Mg:

Molar mass of Mg = 24 g/mol

Moles = Given mass / Molar Mass

Moles of Mg = 4.47 / 24 = 0.18 moles (approx)

Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed

Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3

Unreacted moles of HNO3 = Total Moles - Moles consumed

Unreacted moles of HNO3 = 0.64 moles (approx)

Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate

From the given options, we can see that 0.632 moles is the closest value to our answer

Therefore, 0.632 moles will remain after the reaction

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4 years ago

PH measures the concentration of Hydrogen ions in a substance.<br><br>a. True<br>b False

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3 years ago

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3 years ago

If a 1.271-g sample of aluminum metal is heated in a chlorine gas atmosphere,

Gala2k [10]

Answer:

AlCl₃

Explanation:

Data Given:

Mass of aluminum metal = 1.271 g

Mass of aluminum chloride = 6.280 g

Empirical formula of aluminum chloride = ?

Solution:

First find mass of Chlorine

As 6.280 g of aluminum chloride produced by 1.271 g so

the mass of chlorine in 6.280 g will be 6.280 g -1.271 g)

Mass of chlorine = 5.009 g

Now

Find the number of moles of Al and chlorine in aluminum chloride.

Molar Mass of Al = 26.98 g/mole

Molar mass of Cl = 35.5 g/mol

Mole of Al

Formula Used

no. of moles = mass in grams / molar mass . . . . . . . . (1)

Put values in equation 1

no. of moles of Al = 1.271 g / 26.98 g/ mole

no. of moles of Al = 0.0471

Mole of Chlorine

Formula used

no. of moles = mass in grams / molar mass . . . . . . . . (1)

Put values in equation 1

no. of moles of Cl = 5.009 g / 35.5 g/ mole

no. of moles of Cl = 0.1411

Now we have

Al = 0.0471 moles

Cl = 0.1411 moles

As we Know

Empirical formula shows the simplest ratio of atoms in the molecule but not whole numbers of atoms in a compound.

So,

The ratio of moles of Al to chlorine is

Al : Cl

0.0471 0.1411

Divide the ratio by smallest number to get simplest whole number ratio

Al : Cl

0.0471 / 0.0471 0.1411/ 0.0471

Al : Cl

1 : 3

The simplest ratio of Al to cl is 1 to 3, so the formula will be

Emperical formula of aluminum chloride = AlCl₃

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3 years ago