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bearhunter [10]
4 years ago
7

Tin(II) fluoride (SnF2) is often added to toothpaste as an ingredient to prevent tooth decay. What is the mass of F in grams in

33.7 g of the compound?
Chemistry
1 answer:
Alexeev081 [22]4 years ago
6 0

Answer:

In 33.7 grams SnF2 we have 8.17 grams of F

Explanation:

Step 1: Data given

Mass of SnF2 = 33.7 grams

Molar mass of SnF2 = 156.69 g/mol

Molar mass of F = 19.00 g/mol

Step 2: Calculate moles of SnF2

Moles SnF2 = mass / molar mass

Moles SnF2 = 33.7 grams / 156.69 g/mol

Moles SnF2 = 0.215 moles

Step 3: Calculate moles F

For 1 mol SnF2 we have 2 moles F

For 0.215 moles SnF2 we have 2*0.215 = 0.430 moles F

Step 4: Calculate mass F

Mass F = moles F * molar mass F

Mass F = 0.430 moles * 19.00 g/mol

Mass F = 8.17 grams

In 33.7 grams SnF2 we have 8.17 grams of F

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If you started with 35.0 grams of H2S and 40.0 grams of O2, how many grams of S8 would be produced, assuming 95 % yield? 8H2S(g)
Vesnalui [34]
Molar mass

H2S = 34.0 g/mol
O2 = 31.99 g/mol
S8 = 256.52 g/mol

Identifying excess reagent and the limiting of the reaction :

8 H2S(g) + 4 O2(g) = S8(I) +  8 H2O(g)

8 x 34 g H2S -------->  256. 52 g S8
35.0 g ----------------> ??

35.0 x 256.52 / 8 x 34 =

8978.2 / 272 => 33.00 g of S8  

H2S is  the  limiting reactant
---------------------------------------------

4 x 31.99 g O2 --------------- 256.52 g S8
40.0 g O2 --------------------- ??

40.0 x 256.52 / 4 x 31.99 =

10260.8 / 127.96 = 80.16 g of S8  

O2 is the excess reagent is the excess <span>reagent
</span>
------------------------------------------------------------

H2S is the limiting reactant, one that is fully consumed, it is he who determines the mass of S8 produced 

33.0 g ----------- 100%
?? g ------------- 95 %

95 x 33.00 / 100 => 31.35 g

hope this helps!


4 0
3 years ago
Phosphorous trichloride (PCl3) is produced
vladimir1956 [14]

The percentage yield obtained from the given reaction above is 74.8%

<h3>Balanced equation </h3>

P₄ + 6Cl₂ → 4PCl₃

Molar mass of P₄ = 31 × 4 = 124 g/mol

Mass of P₄ from the balanced equation = 1 × 124 = 124 g

Molar mass of PCl₃ = 31 + (35.5×3) = 137.5 g/mol

Mass of PCl₃ from the balanced equation = 4 × 137.5 = 550 g

<h3>SUMMARY</h3>

From the balanced equation above,

124 g of P₄ reacted to produce 550 g of PCl₃

<h3>How to determine the theoretical yield </h3>

From the balanced equation above,

124 g of P₄ reacted to produce 550 g of PCl₃

Therefore,

79.12 g of P₄ will react to produce = (79.12  × 550) / 124 = 350.9 g of PCl₃

<h3>How to determine the percentage yield </h3>
  • Actual yield of PCl₃ = 262.6 g
  • Theoretical yield of PCl₃ = 350.9 g
  • Percentage yield =?

Percentage yield = (Actual /Theoretical) × 100

Percentage yield = (262.6 / 350.9) × 100

Percentage yield = 74.8%

Learn more about stoichiometry:

brainly.com/question/14735801

4 0
2 years ago
Cliffs may be affected by weathering. Describe ONE  effect of weathering on a cliff .
Anni [7]
Well if too much weathering occurs, it might break off parts of the cliff and be dangerous to humans, or animals. 
4 0
3 years ago
Read 2 more answers
Chemistry compound empirical formula and identify compounds weightWhich drug is c3 h3 01
valkas [14]

The drug C3H30 is called allenolate / 1-hydroxycyclopropyl

weight is 55.06g/mol

5 0
1 year ago
a 20.5g sample of cleaning detergent contains 8.61g of NH40H.CALCULATE the percentage composition of nitrogen in the cleaning de
AysviL [449]

Answer:

The mass percentage composition of nitrogen in the sample of the cleaning detergent is approximately 16.78%  

Explanation:

The given mass of the sample of the cleaning detergent, m₁ = 20.5 g

The mass of the ammonium hydroxide, NH₄OH in the detergent, m₂ = 8.61 g

The molar mass of NH₄OH = 35.04 g/mol

The molar mass of nitrogen, N = 14.01 g/mol

Therefore, the mass, m₃ of nitrogen, N, in 8.61 g of ammonium hydroxide, NH₄OH, is found as follows;

m₃ = (14.01/35.04) × 8.61 g = (402,087/118,800) g ≈ 3.44 g

The mass of nitrogen, N, in the ammonium hydroxide, NH₄OH, contained in the 20.5 g sample of the cleaning agent, m₃ ≈ 3.44 grams

The percentage composition of nitrogen in the sample of the cleaning detergent, %N is given as follows;

\% Composition = \dfrac{Mass \ of \ component}{Total \ mass \ of \ cleaning \ detergent} \times 100

Therefore;

%N ≈ ((3.44 g)/(20.5 g)) × 100 ≈ 16.78 %

The percentage composition of nitrogen, %N ≈ 16.78%.

6 0
3 years ago
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