how many grams mno2 are needed to generate 126.0 g zn(oh)2? zn + 2mno2 + h2o → zn(oh)2 + mn2o3 mno2 : 86.94…

how many grams mno2 are needed to generate 126.0 g zn(oh)2? zn + 2mno2 + h2o → zn(oh)2 + mn2o3 mno2 : 86.94 g/mol zn(oh)2 : 99.43 g/mol ? g mno2

how many grams mno2 are needed to generate 126.0 g zn(oh)2? zn + 2mno2 + h2o → zn(oh)2 + mn2o3 mno2 : 86.94 g/mol zn(oh)2 : 99.43 g/mol ? g mno2

Answer

Explanation:

Step1: Calculate moles of Zn(OH)₂

$n_{Zn(OH)2}=\frac{m{Zn(OH)2}}{M{Zn(OH)_2}}=\frac{126.0\ g}{99.43\ g/mol}\approx1.27\ mol$

Step2: Determine mole - ratio

From the balanced equation $Zn + 2MnO_2+H_2O\rightarrow Zn(OH)2+Mn_2O_3$, the mole - ratio of $MnO_2$ to $Zn(OH)2$ is $n{MnO_2}:n{Zn(OH)2}=2:1$. So $n{MnO_2} = 2\times n_{Zn(OH)2}$. $n{MnO_2}=2\times1.27\ mol = 2.54\ mol$

Step3: Calculate mass of MnO₂

$m_{MnO_2}=n_{MnO_2}\times M_{MnO_2}=2.54\ mol\times86.94\ g/mol\approx220.8\ g$

Answer:

$220.8\ g$