How many gram of H2S will react with 63.2g of KMnO4 to produce K2SO4 and MnO2? Give answer after rounding off to the nearest integer.

Question

How many gram of H2S will react with 63.2g of KMnO4 to produce K2SO4 and MnO2? Give answer after rounding off to the nearest integer. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

According to the Question: The equivalents of H2S= The equivalents of KMnO4 So, 8× MolesofH2S=3× Moles textofKMnO4 textLetw textis required mass ofH2S. So, (8 × w/(34))=(3 × 63 . 2/(158))w=5g.

Q. How many gram of $H_{2} S$ will react with $63.2 g$ of $K M n O_{4}$ to produce $K_{2} S O_{4}$ and $M n O_{2} ?$ Give answer after rounding off to the nearest integer.

According to the Question:
The equivalents of $H_{2} S =$ The equivalents of $K M n O_{4}$
So,
$8 \times M o l eso f H_{2} S = 3 \times M o l es of K M n O_{4} Let w is required mass of H_{2} S .$
So,
$\frac{8 \times w}{( 34 )} = \frac{3 \times 63.2}{( 158 )} w = 5 g .$

Q. How many gram of H2​S will react with 63.2g of KMnO4​ to produce K2​SO4​ and MnO2​? Give answer after rounding off to the nearest integer.

According to the Question: The equivalents of H2​S= The equivalents of KMnO4​ So, 8×MolesofH2​S=3×MolesofKMnO4​Letwis required mass ofH2​S. So, (34)8×w​=(158)3×63.2​w=5g.

Q. How many gram of $H_{2} S$ will react with $63.2 g$ of $K M n O_{4}$ to produce $K_{2} S O_{4}$ and $M n O_{2} ?$ Give answer after rounding off to the nearest integer.

According to the Question:
The equivalents of $H_{2} S =$ The equivalents of $K M n O_{4}$
So,
$8 \times M o l eso f H_{2} S = 3 \times M o l es of K M n O_{4} Let w is required mass of H_{2} S .$
So,
$\frac{8 \times w}{( 34 )} = \frac{3 \times 63.2}{( 158 )} w = 5 g .$