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Greeley [361]
3 years ago
5

How many moles are in 68 grams of potassium sulfide (K2S)?

Chemistry
1 answer:
luda_lava [24]3 years ago
4 0

Answer:No of moles of potassium sulfide (K2S)=0.61671moles

Explanation:

No of moles is given as   Mass/ molar mass

Here Mass of potassium sulfide (K2S) =68 grams

Molar mass of potassium sulfide (K2S)  = 39.0983 x 2 + 32.065 =110.2616 g/mol

No of moles =68 grams /110.2616 g/mol

=0.61671moles

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Limiting Reactant
liraira [26]

Answer:

(1) Cl₂ is the limiting reactant.

(2) 8.18 g

Explanation:

  • 2Na(s) + Cl₂(g) → 2NaCl(s)

First we <u>convert the given masses of reactants into moles</u>, using their <em>respective molar masses</em>:

  • Na ⇒ 12.0 g ÷ 23 g/mol = 0.522 mol Na
  • Cl₂ ⇒ 5.00 g ÷ 70.9 g/mol = 0.070 mol Cl₂

0.070 moles of Cl₂ would react completely with (2 * 0.070) 0.14 moles of Na. There are more Na moles than that, so Na is the reactant in excess while Cl₂ is the limiting reactant.

Then we <u>calculate how many moles of NaCl are formed</u>, <em>using the limiting reactant</em>:

  • 0.070 mol Cl₂ * \frac{2molNaCl}{1molCl_2} = 0.14 mol NaCl

Finally we <u>convert NaCl moles into grams</u>:

  • 0.14 mol NaCl * 58.44 g/mol = 8.18 g
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HELP ASAP!!! How many grams of lead are in a 41.2 gram sample of lead (II) oxide? With work please :)
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How many total moles of ions are released when the following sample dissolves completely in water? Enter your answer in scientif
Katena32 [7]

<u>Answer:</u> The number of moles of strontium bicarbonate is 7.5\times 10^{-9}mol

<u>Explanation:</u>

Formula units are defined as lowest whole number ratio of ions in an ionic compound. It is calculate by multiplying the number of moles by Avogadro's number which is 6.022\times 10^{23}

We are given:

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So, 4.55\times 10^{15} number of formula units will be contained in = \frac{1}{6.022\times 10^{23}}\times 4.55\times 10^{15}=7.5\times 10^{-9}mol of strontium bicarbonate.

Hence, the number of moles of strontium bicarbonate is 7.5\times 10^{-9}mol

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