1.25 g of a metal (M) reacts with oxygen com | vedclass

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

Question

(B) Mass of metal $(M) = 1.25 \ g$Mass of metal oxide $= 1.68 \ g$Mass of oxygen $= 1.68 \ g - 1.25 \ g = 0.43 \ g$Moles of $M = \frac{1.25 \ g}{69.7

Vedclass · Accepted Answer

$M_2O_3$

Vedclass · Answer

(B) Mass of metal $(M) = 1.25 \ g$Mass of metal oxide $= 1.68 \ g$Mass of oxygen $= 1.68 \ g - 1.25 \ g = 0.43 \ g$Moles of $M = \frac{1.25 \ g}{69.7 \ g \ mol^{-1}} \approx 0.0179 \ mol$Moles of $O = \frac{0.43 \ g}{16.0 \ g \ mol^{-1}} \approx 0.0269 \ mol$Ratio of moles $(M:O) = 0.0179 : 0.0269$Dividing by the smallest value $(0.0179)$: $1 : 1.502 \approx 1 : 1.5$Converting to whole numbers by multiplying by $2$: $2 : 3$$	herefore$ The empirical formula is $M_2O_3$.

$1.25 \ g$ of a metal $(M)$ reacts with oxygen completely to produce $1.68 \ g$ of metal oxide. The empirical formula of the metal oxide is
[molar mass of $M$ and $O$ are $69.7 \ g \ mol^{-1}$ and $16.0 \ g \ mol^{-1}$,respectively]

Similar Questions

Percentage of gold in $14$ carat gold is (in $\%$)

If the weight percentage of $Se$ (at. wt. $78.4$) in peroxidase anhydrous enzyme is $0.5 \%$,what is the minimum molecular weight of the enzyme?

In $CH_3CH_2Br$,the percentage of $Br$ is:

The percentage composition of $H, C$,and $O$ in $H_2C_2O_4$ is respectively:

Complete combustion of $1.80 \, g$ of an oxygen-containing compound $(C_xH_yO_z)$ gave $2.64 \, g$ of $CO_2$ and $1.08 \, g$ of $H_2O$. The percentage of oxygen in the organic compound is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$1.25 \ g$ of a metal $(M)$ reacts with oxygen completely to produce $1.68 \ g$ of metal oxide. The empirical formula of the metal oxide is[molar mass of $M$ and $O$ are $69.7 \ g \ mol^{-1}$ and $16.0 \ g \ mol^{-1}$,respectively]