# Homework 8

- Page ID
- 28865

**Q1.45**

__Question__:

When 12 g of sodium reacts with 23.5 g of chlorine to form sodium chloride, how many grams of sodium chloride is formed? (Assume that sodium chloride is the only product).

__Solution__:

**What we know:** We have a chemical reaction between sodium and chlorine to form only sodium chloride. We are also given the masses of the reactants, 12 g of sodium and 23.5 g of chlorine. We also know that the Law of Conservation of Mass tells us that matter is neither created nor destroyed in a chemical reaction.

**What we're asked for:** The mass, in grams, of sodium chloride formed from the reaction.

**Strategy:**

A. Find the sum of the masses of the reactants from the chemical equation.

**Solution:**

12 g + 23.5 g = 35.5 g

35.5 g of sodium chloride is formed from the reaction.

**Q2.47**

__Question__:

How many Magnesium atoms are there in 4.67 mol of Magnesium?

__Solution__:

**What we know:** We know that we have 4.67 mol of Magnesium.

**What we are asked for:** The number of Magnesium atoms present.

**Strategy:**

A. Familiarize yourself with Avogadro's number which is 6.022 x 10^{23}.

B. Set up a dimensional analysis equation to convert Magnesium moles to Magnesium atoms.

C. Carry out the equation.

**Solution:**

A. 1 mol is equal to 6.022 x 10^{23} of anything, but in this case Magnesium atoms.

B. \[4.67 mol Mg \times \frac{6.022\times 10^{23} atoms Mg}{1 mol Mg}\]

C. The "mol Mg" units cancel so you're left with:

\[4.67 \times (6.022\times 10^{23} atoms Mg) = 2.81\times 10^{24} atoms Mg\]

2.81 x 10^{24} atoms of Magnesium are in 4.67 mol of Magnesium