production of polyvinyl chloride, or pvc. hcl forms when hydrogen gas (h₂) reacts with chlorine gas…

production of polyvinyl chloride, or pvc. hcl forms when hydrogen gas (h₂) reacts with chlorine gas (cl₂).\nh₂ (g) + cl₂ (g) → 2hcl (g)\nwhich expression can be used to determine the mass of cl₂ needed to produce 39.4 grams of hcl?\nthe molar mass of hcl is 36.458 grams per mole, and the molar mass of cl₂ is 70.90 grams per mole. consider molar masses to be exact values.\n39.4 g hcl × 1 mol hcl / 36.458 g hcl × 1 mol cl₂ / 2 mol hcl × 70.90 g cl₂ / 1 mol cl₂\n39.4 g hcl × 1 mol hcl / 36.458 g hcl × 2 mol hcl / 1 mol cl₂ × 70.90 g cl₂ / 1 mol cl₂\n39.4 g hcl × 1 mol hcl / 36.458 g hcl × 1 mol cl₂ / 2 mol hcl × 1 mol cl₂ / 70.90 g cl₂\n39.4 g hcl × 1 mol hcl / 36.458 g hcl × 2 mol hcl / 1 mol cl₂ × 1 mol cl₂ / 70.90 g cl₂\ncalculate the mass of cl₂ needed.\nwrite your answer to the correct number of significant figures. round if necessary.\ngrams

production of polyvinyl chloride, or pvc. hcl forms when hydrogen gas (h₂) reacts with chlorine gas (cl₂).\nh₂ (g) + cl₂ (g) → 2hcl (g)\nwhich expression can be used to determine the mass of cl₂ needed to produce 39.4 grams of hcl?\nthe molar mass of hcl is 36.458 grams per mole, and the molar mass of cl₂ is 70.90 grams per mole. consider molar masses to be exact values.\n39.4 g hcl × 1 mol hcl / 36.458 g hcl × 1 mol cl₂ / 2 mol hcl × 70.90 g cl₂ / 1 mol cl₂\n39.4 g hcl × 1 mol hcl / 36.458 g hcl × 2 mol hcl / 1 mol cl₂ × 70.90 g cl₂ / 1 mol cl₂\n39.4 g hcl × 1 mol hcl / 36.458 g hcl × 1 mol cl₂ / 2 mol hcl × 1 mol cl₂ / 70.90 g cl₂\n39.4 g hcl × 1 mol hcl / 36.458 g hcl × 2 mol hcl / 1 mol cl₂ × 1 mol cl₂ / 70.90 g cl₂\ncalculate the mass of cl₂ needed.\nwrite your answer to the correct number of significant figures. round if necessary.\ngrams

Answer

Explanation:

Step1: Convert mass of HCl to moles

We use the molar - mass of HCl to convert the given mass of HCl to moles. The formula is $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass. So, the number of moles of HCl is $39.4\ g\ HCl\times\frac{1\ mol\ HCl}{36.458\ g\ HCl}$.

Step2: Use mole - ratio from the balanced equation

The balanced chemical equation is $H_2(g)+Cl_2(g)\rightarrow2HCl(g)$. The mole - ratio of $Cl_2$ to $HCl$ is $\frac{1\ mol\ Cl_2}{2\ mol\ HCl}$. So, we multiply the moles of HCl by this mole - ratio: $39.4\ g\ HCl\times\frac{1\ mol\ HCl}{36.458\ g\ HCl}\times\frac{1\ mol\ Cl_2}{2\ mol\ HCl}$.

Step3: Convert moles of $Cl_2$ to mass

We use the molar mass of $Cl_2$ to convert the moles of $Cl_2$ to mass. The molar mass of $Cl_2$ is $70.90\ g/mol$. So, the mass of $Cl_2$ is $39.4\ g\ HCl\times\frac{1\ mol\ HCl}{36.458\ g\ HCl}\times\frac{1\ mol\ Cl_2}{2\ mol\ HCl}\times\frac{70.90\ g\ Cl_2}{1\ mol\ Cl_2}$.

Answer:

First, calculate the expression: [ \begin{align*} &39.4\ g\ HCl\times\frac{1\ mol\ HCl}{36.458\ g\ HCl}\times\frac{1\ mol\ Cl_2}{2\ mol\ HCl}\times\frac{70.90\ g\ Cl_2}{1\ mol\ Cl_2}\ =&\frac{39.4\times1\times1\times70.90}{36.458\times2}\ g\ Cl_2\ =&\frac{39.4\times70.90}{72.916}\ g\ Cl_2\ =&\frac{2793.46}{72.916}\ g\ Cl_2\ \approx&38.3\ g\ Cl_2 \end{align*} ] So the mass of $Cl_2$ needed is $38.3$ grams.