When 12 L of H_2 and 11.2 L of Cl_2 are mix | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The balanced chemical equation is: $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$.According to the stoichiometry,$1 \ volume$ of $H_2$ reacts with

Vedclass · Accepted Answer

$0.8 \ L \ H_2$ and $22.4 \ L \ HCl_{(g)}$

Vedclass · Answer

(C) The balanced chemical equation is: $H_{2(g)} + Cl_{2(g)} ightarrow 2HCl_{(g)}$.According to the stoichiometry,$1 \ volume$ of $H_2$ reacts with $1 \ volume$ of $Cl_2$ to produce $2 \ volumes$ of $HCl_{(g)}$.Given: $V(H_2) = 12 \ L$ and $V(Cl_2) = 11.2 \ L$.Since $Cl_2$ is the limiting reagent $(11.2 \ L $11.2 \ L$ of $Cl_2$ will react with $11.2 \ L$ of $H_2$ to produce $2 	imes 11.2 = 22.4 \ L$ of $HCl_{(g)}$.Remaining $H_2 = 12 \ L - 11.2 \ L = 0.8 \ L$.Thus,the final mixture contains $0.8 \ L$ of $H_2$ and $22.4 \ L$ of $HCl_{(g)}$.

When $12 \ L$ of $H_2$ and $11.2 \ L$ of $Cl_2$ are mixed and allowed to react according to the equation $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$,what is the final composition of the mixture?

Similar Questions

Calculate the amount of carbon dioxide produced when:
$(i)$ $1$ mole of carbon is burnt in air.
$(ii)$ $1$ mole of carbon is burnt in $16 \ g$ of dioxygen.
$(iii)$ $2$ moles of carbon are burnt in $16 \ g$ of dioxygen.

When $10 \, mL$ of $10 \, M \, H_2SO_4$ solution is mixed with $100 \, mL$ of $1 \, M \, NaOH$ solution,the resulting solution will be .......

The set of numerical coefficients that balances the equation $K_2CrO_4 + HCl \to K_2Cr_2O_7 + KCl + H_2O$ is

In the balanced equation of the following reaction, the ratio of $a / b$ is: $aCaCO_3 + bH_3PO_4 \rightarrow pCa_3(PO_4)_2 + qCO_2 + rH_2O$

$100 \, g$ $CaCO_3$ reacts with $1 \, L$ $1 \, N$ $HCl$. On completion of reaction,how much weight of $CO_2$ will be obtained? (in $g$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module