30.0 g of chlorine gas is bubbled through liquid sulfur to produce liquid disulfur dichloride. how much…

Explanation:

Step1: Balance the chemical equation

$2Cl_2 + 2S\rightarrow 2S_2Cl_2$ simplifies to $Cl_2+S\rightarrow S_2Cl_2$

Step2: Calculate molar - masses

The molar mass of $Cl_2$ is $M_{Cl_2}=2\times35.45\ g/mol = 70.90\ g/mol$. The molar mass of $S_2Cl_2$ is $M_{S_2Cl_2}=2\times32.07 + 2\times35.45=64.14 + 70.90 = 135.04\ g/mol$

Step3: Calculate moles of $Cl_2$

The number of moles of $Cl_2$, $n_{Cl_2}=\frac{m_{Cl_2}}{M_{Cl_2}}=\frac{30.0\ g}{70.90\ g/mol}\approx0.423\ mol$

Step4: Determine moles of $S_2Cl_2$

From the balanced chemical equation, the mole - ratio of $Cl_2$ to $S_2Cl_2$ is $1:1$. So, $n_{S_2Cl_2}=n_{Cl_2}=0.423\ mol$

Step5: Calculate mass of $S_2Cl_2$

$m_{S_2Cl_2}=n_{S_2Cl_2}\times M_{S_2Cl_2}=0.423\ mol\times135.04\ g/mol\approx57.0\ g$

Answer:

A. 57.0 grams