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Chemistry

1 answer:

belka [17]3 years ago

3 0

Answer:

Calcium carbonate and citric acid.

Explanation:

Hello!

In this case, considering the text, we can notice that calcium carbonate combines with citric acid to produce the calcium citrate water and carbon dioxide as shown on the following equation:

$3CaCO_3+2C_6H_8O_7\rightarrow Ca_3C_{12}H_{10}O_{14}+CO_2+H_2O$

Whereas the reactants, at the left side, are identified as calcium carbonate and citric acid.

Best regards!

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2AlF3 + 3K2O → 6KF + Al2O3

Agata [3.3K]

<h3>Answer:</h3>

7.4797 g AlF₃

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Reading a Periodic Table
Using Dimensional Analysis

<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] 6 mol KF = 2 mol AlF₃

Molar Mass of K - 39.10 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of Al - 26.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.1 g/mol

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

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.1 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})$
Multiply/Divide: $\displaystyle 7.47966 \ g \ AlF_3$