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Alex777 [14]
3 years ago
14

How many moles of BaSO4 are formed if 0.5 moles of Na2 SO4 react with 60 g of BaCl2? Na2 SO4 + BaCl2 → BaSO4 + 2NaCl

Chemistry
2 answers:
faust18 [17]3 years ago
6 0
<span>the balanced equation for the reaction is as follows
Na</span>₂<span>SO</span>₄<span> + BaCl</span>₂<span> ----> 2NaCl + BaSO</span>₄ 
<span>stoichiometry of Na</span>₂<span>SO</span>₄<span> to BaCl</span>₂<span> is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.
                                                                           
number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other
                                                                         
therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl</span>₂<span> to BaSO</span>₄<span> is 1:1.
                                                                           
therefore number of BaSO4 moles formed - 0.288 mol</span>
ozzi3 years ago
6 0

Answer: 0.29

Explanation:

Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}

Putting values in above equation, we get:

\text{Moles of}BaCl_2}=\frac{60g}{208g/mol}=0.29mol

Moles of Na_2SO_4 = 0.5 moles

According to stochiometry, 1 mole of BaCl_2 reacts with 1 mole of Na_2SO_4

Thus 0.29 moles of BaCl_2 will react with 0.29 moles of Na_2SO_4

Thus BaCl_2 is the limiting reagent as it limits the formation of products and Na_2SO_4 is the excess reagent.

1 mole of BaCl_2 produces = 1 mole of BaSO_4

0.29 moles of BaCl_2 will produce= 0.29 moles of BaSO_4

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6 0
2 years ago
11.9 g Cl2 is reacted with 10.7 g NaOH. How many moles of NaCl are produced?
melisa1 [442]

Answer:

1. 7.256g of NaCl

2. 47.33g of Cl2

Explanation:

2 moles of Na reacts to produce 2 moles of NaCl

8 moles of Na will still produce 8 moles of NaCl

Mass of NaCl = molar mass of Nacl/moles of Nacl

=58.5/8

=7.256g of NaCl

From the equation, 2 moles of Na reacts with 1 mole of Cl2

3/2 moles of Cl2 will react with 3 moles of Na

Mass of Cl2 = 71/1.5

=47.33g of Cl2

Explanation:

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5 0
3 years ago
How many particles are in 47.7 g of Magnesium? (Round the average
Lynna [10]

Answer:

1.18 × 10²⁴ particles Mg

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

47.7 g Mg

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of Mg - 24.31 g/mol

<u>Step 3: Convert</u>

<u />47.7 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg} )(\frac{6.022 \cdot 10^{23} \ particles \ Mg}{1 \ mol \ Mg} ) = 1.18161 × 10²⁴ particles Mg

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

1.18161 × 10²⁴ particles Mg ≈ 1.18 × 10²⁴ particles Mg

3 0
2 years ago
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