Three substances A,B and C react to form E an | vedclass

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

Question

(C) The balanced chemical equation is $2A + 3B + C \rightarrow 4D + 2E$.First,calculate the mass of one stoichiometric unit of the reactants: $2(40) +

Vedclass · Accepted Answer

$390$

Vedclass · Answer

(C) The balanced chemical equation is $2A + 3B + C ightarrow 4D + 2E$.First,calculate the mass of one stoichiometric unit of the reactants: $2(40) + 3(30) + 1(20) = 80 + 90 + 20 = 190 \ g$.Since $570 \ g$ of the mixture is used,the number of stoichiometric units reacting is $570 / 190 = 3$.According to the stoichiometry,$1$ unit of reaction produces $2$ moles of $E$. Therefore,$3$ units produce $3 	imes 2 = 6 \ moles$ of $E$.Using the law of conservation of mass: $Mass_{reactants} = Mass_{products}$.$570 \ g = Mass_{D} + Mass_{E}$.Mass of $D$ produced $= 3 	imes (4 	imes 15) = 3 	imes 60 = 180 \ g$.Mass of $E = 570 \ g - 180 \ g = 390 \ g$.

Similar Questions

At $STP$,the volume of $4.4 \, g$ of an unknown gas is $2.24 \, L$. Which of the following is this gas?

$A$ cylinder contains a mixture of $5 \ g$ of $N_2$ and $6 \ g$ of $Ar$ gases. If the total pressure of the mixture of the gases in the cylinder is $30 \ bar$,then the partial pressure of $N_2$ gas is (in $bar$)

$A$ $100 \ mL$ solution of $CH_3CH_2MgBr$ on treatment with methanol produces $2.24 \ mL$ of a gas at $STP$. The weight of gas produced is $......mg$. [nearest integer]

The minimum number of moles of $O_2$ required for complete combustion of $1$ mole of propane and $2$ moles of butane is...

$A$ $100 \ mL$ solution was made by adding $1.43 \ g$ of $Na_2CO_3 \cdot xH_2O$. The normality of the solution is $0.1 \ N$. The value of $x$ is .......... (The atomic mass of $Na$ is $23 \ g/mol$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module