Question

1. $c_2h_6$ combusts with $o_2$ to produce $h_2o$ and $co_2$. if i have 48 grams of $c_2h_6$, how many liters of water would be produced? $2c_2h_6 + 7o_2 \

ightarrow 6h_2o + 4co_2$ 8. if 67.8 g of $cu$ is reacted with $agno_3$, how many liters of $cu(no_3)_2$ are produced? $cu(s) + 2agno_{3(aq)} \
ightarrow cu(no_3)_{2(aq)} + 2ag(s)$ (and another problem about reaction with $mg$ to produce $mgcl_2$ and $h_2$ gas, with 56.2 g of $hcl$ asking for moles of $mgcl_2$)

Explanation:

For Question 7:

Step1: Calculate molar mass of $\text{C}_2\text{H}_6$

Molar mass of $\text{C}_2\text{H}_6 = 2\times12.01 + 6\times1.008 = 30.07\ \text{g/mol}$

Step2: Find moles of $\text{C}_2\text{H}_6$

$\text{Moles of }\text{C}_2\text{H}_6 = \frac{48\ \text{g}}{30.07\ \text{g/mol}} \approx 1.596\ \text{mol}$

Step3: Mole ratio of $\text{C}_2\text{H}_6:\text{H}_2\text{O}$

From balanced equation: $2:6 = 1:3$, so $\text{Moles of }\text{H}_2\text{O} = 1.596\ \text{mol} \times 3 = 4.788\ \text{mol}$

Step4: Convert moles to volume (STP)

At STP, 1 mol gas = 22.4 L. $\text{Volume of }\text{H}_2\text{O} = 4.788\ \text{mol} \times 22.4\ \text{L/mol} \approx 107.25\ \text{L}$

Step1: Calculate molar mass of $\text{Cu}$

Molar mass of $\text{Cu} = 63.55\ \text{g/mol}$

Step2: Find moles of $\text{Cu}$

$\text{Moles of }\text{Cu} = \frac{67.8\ \text{g}}{63.55\ \text{g/mol}} \approx 1.067\ \text{mol}$

Step3: Mole ratio of $\text{Cu}:\text{Cu(NO}_3\text{)}_2$

From balanced equation: $1:1$, so $\text{Moles of }\text{Cu(NO}_3\text{)}_2 = 1.067\ \text{mol}$

Step4: Convert moles to volume (STP)

At STP, 1 mol gas = 22.4 L. $\text{Volume of }\text{Cu(NO}_3\text{)}_2 = 1.067\ \text{mol} \times 22.4\ \text{L/mol} \approx 23.90\ \text{L}$

Step1: Calculate molar mass of $\text{HCl}$

Molar mass of $\text{HCl} = 1.008 + 35.45 = 36.458\ \text{g/mol}$

Step2: Find moles of $\text{HCl}$

$\text{Moles of }\text{HCl} = \frac{56.2\ \text{g}}{36.458\ \text{g/mol}} \approx 1.542\ \text{mol}$

Step3: Mole ratio of $\text{HCl}:\text{MgCl}_2$

From balanced equation: $2:1$, so $\text{Moles of }\text{MgCl}_2 = \frac{1.542\ \text{mol}}{2} = 0.771\ \text{mol}$

Answer:

$\approx 107.3\ \text{L}$

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For Question 8: