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ratelena [41]
3 years ago
15

How many Mg atoms are present in 170g of Mg

Chemistry
1 answer:
PIT_PIT [208]3 years ago
8 0
In order to calculate the number of atoms, we must first know the number of moles present. And
moles = (mass present) / (molecular mass)
Therefore, the moles of Mg present are
170 / 24 = 7.08
The number of atoms in a mole of substance is given by Avagadro's Number which is 6.02 x 10^23
Since there are 7.08 moles, there are:
7.08 * 6.02*10^23
= 4.26 * 10^24 atoms
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Calculate the number of hydrogen atoms in 33.0 g CH4
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Answer:

The number of hydrogen atoms is 4.96x10²⁴.

 

Explanation:

The number of atoms can be found with the following equation:

n = N*\eta_{H}

Where:

N: is the Avogadro's number = 6.022x10²³ atoms/mol

η: is the number of moles of hydrogen

n: is the number of hydrogen atoms

First, we need to find the number of hydrogen moles. The number of moles of CH₄ is:

\eta_{CH_{4}} = \frac{m}{M}

Where:

m: is the mass of methane = 33 g

M: is the molar mass of methane = 16.04 g/mol                      

\eta_{CH_{4}} = \frac{33 g}{16.04 g/mol} = 2.06 mol

Now, since we have 4 hydrogen atoms in 1 mol of methane, the number of moles of hydrogen is:

\eta_{H} = 2.06\: mol\: CH_{4}*4 \frac{mol\: H}{mol \: CH_{4}} = 8.24 mol

Hence, the number of hydrogen atoms is:

n = N*\eta_{H} = 6.022 \cdot 10^{23} \: atoms/mol*8.24 mol = 4.96 \cdot 10^{24} atoms

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When 4.15 grams of silver nitrate is reacted with 1.11 grams of iron(III) chloride, which best represents the amount of silver c
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Answer:

The mass of silver chloride produced = 2.202 g

Explanation:

Equation of the reaction is given below

3AgNO₃(aq) + FeCl₃(aq) ----> 3AgCl(s) + Fe(NO₃)₃(aq)

molar mass of AgNO₃ = 170 g/mol

molar mass of FeCl₃ = 233.5 g/mol

molar mass of AgCl = 143.5 g/mol

3 moles of silver nitrate reacts with 1 mole of iron (iii) chloride to give 3 moles of silver nitrate

4.15 grams of AgNO₃ = 4.15/170 = 0.0244 moles of AgNO₃

1.11 grams of FeCl₃ = 1.11/233.5 = 0.0047 moles of FeCl₃

mole ratio of AgNO₃ to FeCl₃ = 0.0244/0.0047 = 5 : 1

therefore, FeCl₃ is the limiting reactant

0.0047 moles of FeCl₃ reacting will produce 0.0047 *  3 moles of AgCl = 0.0141 moles of AgCl

0.0141 moles of AgCl = 0.0141 * 143.5 g of AgCl = 2.02 g of AgCl =

Therefore mass of silver chloride produced = 2.202 g

3 0
3 years ago
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