Question

consider the following reaction: ch₄(g) + 4cl₂(g) → ccl₄(g) + 4hcl(g)
what mass of ccl₄ will be formed if 1.20 moles of methane react with 1.11 moles of chlorine?
185 g
683 g
42.7 g
19.7 g
171 g
question 5
1 pts
equal masses (in grams) of hydrogen gas and oxygen gas are reacted to form water. which substance is limiting?
hydrogen gas is limiting.
more information is needed to answer this question.
water is limiting.
oxygen gas is limiting.
nothing is limiting.

Explanation:

Step1: Identify the limiting reactant

The balanced - equation is $\mathrm{CH}_{4}(g)+4\mathrm{Cl}_{2}(g)
ightarrow\mathrm{CCl}_{4}(g) + 4\mathrm{HCl}(g)$. The mole - ratio of $\mathrm{CH}_{4}$ to $\mathrm{Cl}_{2}$ is $1:4$. Given $n_{\mathrm{CH}_{4}} = 1.20\ mol$ and $n_{\mathrm{Cl}_{2}}=1.11\ mol$. For $1.20\ mol$ of $\mathrm{CH}_{4}$, the required amount of $\mathrm{Cl}_{2}$ is $n_{\mathrm{Cl}_{2}\text{(required)}}=1.20\ mol\times4 = 4.80\ mol$. Since $1.11\ mol$ of $\mathrm{Cl}_{2}$ is available and $4.80\ mol$ is required for complete reaction of $\mathrm{CH}_{4}$, $\mathrm{Cl}_{2}$ is the limiting reactant.

Step2: Calculate moles of $\mathrm{CCl}_{4}$ formed

From the balanced equation, the mole - ratio of $\mathrm{Cl}_{2}$ to $\mathrm{CCl}_{4}$ is $4:1$. So, $n_{\mathrm{CCl}_{4}}=\frac{1}{4}n_{\mathrm{Cl}_{2}}$. Substituting $n_{\mathrm{Cl}_{2}} = 1.11\ mol$, we get $n_{\mathrm{CCl}_{4}}=\frac{1.11\ mol}{4}=0.2775\ mol$.

Step3: Calculate the mass of $\mathrm{CCl}_{4}$

The molar mass of $\mathrm{CCl}_{4}$ is $M_{\mathrm{CCl}_{4}}=1\times12.01\ g/mol + 4\times35.45\ g/mol=153.81\ g/mol$. Using the formula $m = n\times M$, we have $m_{\mathrm{CCl}_{4}}=n_{\mathrm{CCl}_{4}}\times M_{\mathrm{CCl}_{4}}=0.2775\ mol\times153.81\ g/mol\approx42.7\ g$.

for Question 5:

Step1: Write the balanced chemical equation

The balanced equation for the reaction between hydrogen gas and oxygen gas to form water is $2\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)
ightarrow2\mathrm{H}_{2}\mathrm{O}(l)$. The mole - ratio of $\mathrm{H}_{2}$ to $\mathrm{O}_{2}$ is $2:1$.

Step2: Let the mass of both $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ be $m\ g$

The molar mass of $\mathrm{H}_{2}$ is $M_{\mathrm{H}_{2}} = 2.02\ g/mol$, so the number of moles of $\mathrm{H}_{2}$ is $n_{\mathrm{H}_{2}}=\frac{m\ g}{2.02\ g/mol}$. The molar mass of $\mathrm{O}_{2}$ is $M_{\mathrm{O}_{2}} = 32.00\ g/mol$, so the number of moles of $\mathrm{O}_{2}$ is $n_{\mathrm{O}_{2}}=\frac{m\ g}{32.00\ g/mol}$.

Step3: Determine the limiting reactant

From the mole - ratio, for every 1 mole of $\mathrm{O}_{2}$, 2 moles of $\mathrm{H}_{2}$ are required. The number of moles of $\mathrm{H}_{2}$ required to react with $n_{\mathrm{O}_{2}}$ moles of $\mathrm{O}_{2}$ is $n_{\mathrm{H}_{2}\text{(required)}} = 2n_{\mathrm{O}_{2}}$. Substituting $n_{\mathrm{O}_{2}}=\frac{m}{32.00}\ mol$, we get $n_{\mathrm{H}_{2}\text{(required)}}=\frac{2m}{32.00}\ mol$. And $n_{\mathrm{H}_{2}}=\frac{m}{2.02}\ mol$. Since $\frac{m}{2.02}\ mol>\frac{2m}{32.00}\ mol$, $\mathrm{O}_{2}$ is the limiting reactant.

Answer:

$42.7\ g$