Question

1. what is the molarity of a solution which contains 58.5 g of sodium chloride dissolved in 0.500 l of solution?

1. zn(s) + 2hcl(aq) → h₂(g) + zncl₂(aq)

when 25.0 g of zn reacts, how many l of h₂ gas are formed at stp?

1. complete the unbalanced double replacement equation and indicate if a precipitate forms.

naoh(aq) + mgcl₂(aq) → ______________

1. balance the equation:

nh₃(g) + o₂(g) → no₂(g) + h₂o(g)

Explanation:

Step1: Calculate moles of NaCl

Molar mass of NaCl: $M(\text{NaCl}) = 23.0\ \text{g/mol} + 35.5\ \text{g/mol} = 58.5\ \text{g/mol}$
Moles: $n(\text{NaCl}) = \frac{m}{M} = \frac{58.5\ \text{g}}{58.5\ \text{g/mol}} = 1.00\ \text{mol}$

Step2: Calculate molarity of solution

Molarity formula: $M = \frac{n}{V}$
$M = \frac{1.00\ \text{mol}}{0.500\ \text{L}} = 2.00\ \text{mol/L}$

Step1: Calculate moles of Zn

Molar mass of Zn: $M(\text{Zn}) = 65.4\ \text{g/mol}$
Moles: $n(\text{Zn}) = \frac{25.0\ \text{g}}{65.4\ \text{g/mol}} \approx 0.382\ \text{mol}$

Step2: Relate moles of Zn to H₂

From reaction: $n(\text{H}_2) = n(\text{Zn}) \approx 0.382\ \text{mol}$

Step3: Find volume at STP

At STP, 1 mol gas = 22.4 L
$V(\text{H}_2) = 0.382\ \text{mol} \times 22.4\ \text{L/mol} \approx 8.56\ \text{L}$

Step1: Predict double replacement products

Swap cations: $\text{NaOH} + \text{MgCl}_2
ightarrow \text{Mg(OH)}_2 + \text{NaCl}$

Step2: Balance the equation

Balance atoms: $2\text{NaOH(aq)} + \text{MgCl}_2\text{(aq)}
ightarrow \text{Mg(OH)}_2\text{(s)} + 2\text{NaCl(aq)}$

Step3: Identify precipitate

$\text{Mg(OH)}_2$ is insoluble in water.

Step1: Balance N atoms

Add coefficient 4 to $\text{NH}_3$ and $\text{NO}_2$:
$4\text{NH}_3\text{(g)} + \text{O}_2\text{(g)}
ightarrow 4\text{NO}_2\text{(g)} + \text{H}_2\text{O(g)}$

Step2: Balance H atoms

Add coefficient 6 to $\text{H}_2\text{O}$:
$4\text{NH}_3\text{(g)} + \text{O}_2\text{(g)}
ightarrow 4\text{NO}_2\text{(g)} + 6\text{H}_2\text{O(g)}$

Step3: Balance O atoms

Total O on right: $4\times2 + 6\times1 = 14$, so add coefficient 7 to $\text{O}_2$:
$4\text{NH}_3\text{(g)} + 7\text{O}_2\text{(g)}
ightarrow 4\text{NO}_2\text{(g)} + 6\text{H}_2\text{O(g)}$

Answer:

$2.00\ \text{mol/L}$

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