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poizon [28]
3 years ago
11

How many moles of Ag will be produced from 16g of Cu, assuming AgNO3 is available in excess

Chemistry
1 answer:
andreyandreev [35.5K]3 years ago
3 0

Answer:

Assuming the reaction you are referring to is the following:

AgNO3 + Cu --> Cu(NO3)2 + Ag

First, we should balance the reaction:

2 AgNO3 + Cu --> Cu(NO3)2 + 2 Ag

Now we carry out dimensional analysis, taking into account the molar ratios of different species as given in the balanced reaction and molar masses (which we can look up in the periodic table).

We begin with what we know, which is that we have 16.0 g Cu reacting and work our way toward the quantity that we would like to calculate, namely moles of Ag.

16.0 g Cu x   1 mol Cu    x   2 mol Ag    = 0.504 mol Ag

                     63.5 g Cu       1 mol Cu

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How many atoms are there in 7.80 moles of germanium
Orlov [11]
<h3>Answer:</h3>

4.70 × 10²⁴ atoms Ge

<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>Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

7.80 mol Ge

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

Avogadro's Number

<u>Step 3: Convert</u>

\displaystyle 7.80 \ mol \ Ge(\frac{6.022 \cdot 10^{23} \ atoms \ Ge}{1 \ mol \ Ge} ) = 4.69716 × 10²⁴ atoms Ge

<u>Step 4: Check</u>

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

4.69716 × 10²⁴ atoms Ge ≈ 4.70 × 10²⁴ atoms Ge

8 0
3 years ago
What mass (g) of magnesium nitride (Mg3N2) can be made from the reaction of 1.22 g of magnesium with excess nitrogen? __Mg + __N
Citrus2011 [14]
<h3>Answer:</h3>

1.69 g Mg₃N₂

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

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

[RxN] 3 mol Mg → Mg₃N₂

[PT] Molar Mass of Mg - 24.31 g/mol

[PT] Molar Mass of N - 14.01 g/mol

Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                      \displaystyle 1.22 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{1 \ mol \ Mg_3N_2}{3 \ mol \ Mg})(\frac{100.95 \ g \ Mg_3N_2}{1 \ mol\ Mg_3N_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 1.68873 \ g \ Mg_3N_2

<u>Step 4: Check</u>

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

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂

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