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erica [24]
3 years ago
5

2AlF3 + 3K2O → 6KF + Al2O3 How many grams of AlF3would it take to make 15.524 g of KF?

Chemistry
2 answers:
attashe74 [19]3 years ago
5 0

Answer:

7.4798

Explanation:

i just did the quiz, the guy above me is wrong.

mr_godi [17]3 years ago
3 0
<h3>Answer:</h3>

7.4797 g AlF₃

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃

[Given] 15.524 g KF

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

[RxN] 2 mol AlF₃ = 6 mol KF

Molar Mass of Al - 26.98 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of K - 39.10 g/mol

Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.10 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                              \displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.10 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})
  2. Multiply/Divide:                                                                                                    \displaystyle 7.47966 \ g \ AlF_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

7.47966 g AlF₃ ≈ 7.4797 g AlF₃

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M(HNO3)*V(HNO3) = M(LiOH)*M(LiOH)

M(HNO3)*50.0mL = 0.100M*36.90 mL

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Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou
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Answer:

4.36~g~XY

Explanation:

In this case, we can start with the reaction:

2X + Y_2~->~2XY

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:

3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X

4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2

Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):

\frac{0.04~mol~X}{1}=~0.04

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0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol Y_2 = 48 g Y_2 (therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY

I hope it helps!

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3 years ago
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Hey there!

To calculate the percent by mass of the Ca(NO₃)₂ we need to find the total mass first by adding.

896.92 + 22.63 = 919.55

In total, the solution is 919.55 grams.

To find the percent of Ca(NO₃)₂ in the solution, divide the mass of Ca(NO₃)₂ by the total mass and multiply by 100.

22.63 ÷ 919.55 = 0.0246

0.0246 x 100 = 2.46

Ca(NO₃)₂ makes up 2.46% of the solution.

Hope this helps!

6 0
3 years ago
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