
Which is formed when ${K_2}C{r_2}{O_7}$, $CaC{l_2}$ and $conc.{H_2}S{O_4}$ are heated?
A. $C{r_2}{(S{O_4})_3}$
B. $CrC{l_3}$
C. $Cr{O_2}C{l_2}$
D. ${K_2}Cr{O_4}$
Answer
570.9k+ views
Hint: We know that ${K_2}C{r_2}{O_7}$ is known as potassium dichromate. $CaC{l_2}$ is known as calcium chloride and ${H_2}S{O_4}$ is known as sulphuric acid. Sulphuric acid is a very strong acid. Sulphuric acid when exposed to human skin can cause severe burns.
Complete step by step solution:
We know that ${K_2}C{r_2}{O_7}$ is a very important inorganic compound. Potassium dichromate is a very strong oxidising agent. Whenever ${K_2}C{r_2}{O_7},CaC{l_2}$ and $conc.{H_2}S{O_4}$ are heated, we observe a very interesting chemical reaction. The chemical equation can be written as $CaC{l_2} + {H_2}S{O_4} + {K_2}C{r_2}{O_7}\xrightarrow{\Delta }Cr{O_2}C{l_2} + KHS{O_4} + CaS{O_4} + {H_2}O$. We can see that it forms four products including water. If we try to balance this chemical equation we will get
$2CaC{l_2} + 4{H_2}S{O_4} + {K_2}C{r_2}{O_7}\xrightarrow{\Delta }2Cr{O_2}C{l_2} + 2KHS{O_4} + 2CaS{O_4} + 3{H_2}O$.
We can see that the products formed are $Cr{O_2}C{l_2},KHS{O_4},CaS{O_4},{H_2}O$.
So from the above explanation and discussion it is clear to us that the correct answer of the given question is option C. $Cr{O_2}C{l_2}$
Hence, the correct answer is option C.
Additional information:
Sulphuric acid is odourless, colourless and a very strong acid. It is soluble in water. The reaction involving sulphuric acids is an exothermic reaction because it releases heat. Potassium dichromate is also an odourless compound. The colour of potassium dichromate is red-orange. It is used to oxide other inorganic and organic compounds. Calcium chloride is an important inorganic compound. It is formed when hydrochloric acid reacts with calcium hydroxide. It is a white coloured crystalline solid. It is highly soluble in water.
Note: Always use sulphuric acid carefully while performing experiments because it can cause severe burns. Always remember that when potassium dichromate and calcium chloride is heated in the presence of sulphuric acid we get this reaction - $2CaC{l_2} + 4{H_2}S{O_4} + {K_2}C{r_2}{O_7}\xrightarrow{\Delta }2Cr{O_2}C{l_2} + 2KHS{O_4} + 2CaS{O_4} + 3{H_2}O$
Complete step by step solution:
We know that ${K_2}C{r_2}{O_7}$ is a very important inorganic compound. Potassium dichromate is a very strong oxidising agent. Whenever ${K_2}C{r_2}{O_7},CaC{l_2}$ and $conc.{H_2}S{O_4}$ are heated, we observe a very interesting chemical reaction. The chemical equation can be written as $CaC{l_2} + {H_2}S{O_4} + {K_2}C{r_2}{O_7}\xrightarrow{\Delta }Cr{O_2}C{l_2} + KHS{O_4} + CaS{O_4} + {H_2}O$. We can see that it forms four products including water. If we try to balance this chemical equation we will get
$2CaC{l_2} + 4{H_2}S{O_4} + {K_2}C{r_2}{O_7}\xrightarrow{\Delta }2Cr{O_2}C{l_2} + 2KHS{O_4} + 2CaS{O_4} + 3{H_2}O$.
We can see that the products formed are $Cr{O_2}C{l_2},KHS{O_4},CaS{O_4},{H_2}O$.
So from the above explanation and discussion it is clear to us that the correct answer of the given question is option C. $Cr{O_2}C{l_2}$
Hence, the correct answer is option C.
Additional information:
Sulphuric acid is odourless, colourless and a very strong acid. It is soluble in water. The reaction involving sulphuric acids is an exothermic reaction because it releases heat. Potassium dichromate is also an odourless compound. The colour of potassium dichromate is red-orange. It is used to oxide other inorganic and organic compounds. Calcium chloride is an important inorganic compound. It is formed when hydrochloric acid reacts with calcium hydroxide. It is a white coloured crystalline solid. It is highly soluble in water.
Note: Always use sulphuric acid carefully while performing experiments because it can cause severe burns. Always remember that when potassium dichromate and calcium chloride is heated in the presence of sulphuric acid we get this reaction - $2CaC{l_2} + 4{H_2}S{O_4} + {K_2}C{r_2}{O_7}\xrightarrow{\Delta }2Cr{O_2}C{l_2} + 2KHS{O_4} + 2CaS{O_4} + 3{H_2}O$
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