Question

unit 6 stoichiometry
name:
show all work for credit including labeling the equation with the given and unknown. box your final
answer, including units. round answer to 2 places after the decimal.
1 mole = molar mass (calculate from the periodic table)
1 mole = 22.4 l

1. how many moles of no gas can combine with 527 moles of oxygen gas?

\t\t\t\t\t2no$_{(g)}$ + o$_{2(g)}$ $
ightarrow$ 2no$_{2(g)}$

1. how many moles of carbon dioxide are formed when 40.0 moles of oxygen gas are consumed?

\t\t\t\t\tch$_{4(g)}$ + 2o$_{2(g)}$ $longrightarrow$ co$_{2(g)}$ + 2h$_2$o$_{(g)}$

1. calculate the number of moles of hydrogen gas produced by the reaction of 14.5 grams of

potassium metal.
\t\t\t\t\t2k$_{(s)}$ + 2h$_2$o$_{(l)}$ $longrightarrow$ 2koh$_{(aq)}$ + h$_2$(g)

1. how many liters of co$_2$ can be produced from 14.5 l of c$_2$h$_2$?

\t\t\t\t\t2c$_2$h$_{2(g)}$ + 5o$_{2(g)}$ $longrightarrow$ 4co$_{2(g)}$ + 2h$_2$o$_{(g)}$

Explanation:

Step1: Molar ratio identification

From $2NO_{(g)} + O_{2(g)}
ightarrow 2NO_{2(g)}$, the mole ratio of $NO$ to $O_2$ is $\frac{2\ \text{mol }NO}{1\ \text{mol }O_2}$

Step2: Calculate moles of NO

$\text{Moles of }NO = 527\ \text{mol }O_2 \times \frac{2\ \text{mol }NO}{1\ \text{mol }O_2} = 1054.00\ \text{mol}$

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Step1: Molar ratio identification

From $CH_{4(g)} + 2O_{2(g)}
ightarrow CO_{2(g)} + 2H_2O_{(g)}$, the mole ratio of $CO_2$ to $O_2$ is $\frac{1\ \text{mol }CO_2}{2\ \text{mol }O_2}$

Step2: Calculate moles of $CO_2$

$\text{Moles of }CO_2 = 40.0\ \text{mol }O_2 \times \frac{1\ \text{mol }CO_2}{2\ \text{mol }O_2} = 20.00\ \text{mol}$

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Step1: Find molar mass of K

From periodic table, molar mass of K = $39.10\ \text{g/mol}$

Step2: Calculate moles of K

$\text{Moles of }K = \frac{14.5\ \text{g}}{39.10\ \text{g/mol}} \approx 0.3708\ \text{mol}$

Step3: Molar ratio identification

From $2K_{(s)} + 2H_2O_{(l)}
ightarrow 2KOH_{(aq)} + H_{2(g)}$, the mole ratio of $H_2$ to $K$ is $\frac{1\ \text{mol }H_2}{2\ \text{mol }K}$

Step4: Calculate moles of $H_2$

$\text{Moles of }H_2 = 0.3708\ \text{mol }K \times \frac{1\ \text{mol }H_2}{2\ \text{mol }K} \approx 0.19\ \text{mol}$

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Step1: Molar ratio identification

From $2C_2H_{2(g)} + 5O_{2(g)}
ightarrow 4CO_{2(g)} + 2H_2O_{(g)}$, the volume ratio of $CO_2$ to $C_2H_2$ is $\frac{4\ \text{L }CO_2}{2\ \text{L }C_2H_2}$

Step2: Calculate volume of $CO_2$

$\text{Volume of }CO_2 = 14.5\ \text{L }C_2H_2 \times \frac{4\ \text{L }CO_2}{2\ \text{L }C_2H_2} = 29.00\ \text{L}$

Answer:

1. $\boldsymbol{1054.00}$ moles of NO
2. $\boldsymbol{20.00}$ moles of $CO_2$
3. $\boldsymbol{0.19}$ moles of $H_2$
4. $\boldsymbol{29.00}$ liters of $CO_2$