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inn [45]
3 years ago
7

Which equation shows how to calculate how many grams (g) of Mg(OH)2 would be produced from 4mol KOH? The balanced reaction is:

Chemistry
1 answer:
Simora [160]3 years ago
6 0

Answer:

4 mol of KOH would produce 116.6 g of Mg(OH)₂

Explanation:

According to the following balanced equation:

  • MgCl₂+ 2 KOH -----> Mg(OH)₂ + 2 KCL

One can note that 2 mol of KOH react with MgCl₂ to produce 1 mol of Mg(OH)₂.

using cross multiplication  

2 mol of KOH → 1 mol of Mg(OH)₂.

4 mol of KOH → ?? mol of Mg(OH)₂.

no of moles of  Mg(OH)₂ = (1 mol* 4 mol) / 2 mol =2 mol

Now we can convert moles of  Mg(OH)₂ to grams using the formula

mass of Mg(OH)₂= (no. of moles * molar mass) = (2 mol * 58.3g/mol) = 116.6 g

  • So, 4 mol of KOH would produce 116.6 g of Mg(OH)₂.
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510 g NO₂

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  • Reading the Periodic Table
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  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)

<u>Step 2: Define conversions</u>

Avogadro's Number

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<u>Step 3: Use Dimensional Analysis</u>

<u />6.7 \cdot 10^{24} \ molecules \ NO_2(\frac{1 \ mol \ NO_2}{6.022 \cdot 10^{23} \ molecules \ NO_2} )(\frac{46.01 \ g \ NO_2}{1 \ mol \ NO_2} ) = 511.901 g NO₂

<u>Step 4: Check</u>

<em>We are given 2 sig figs. Follow sig fig rules.</em>

511.901 g NO₂ ≈ 510 g NO₂

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