How many grams of MnO-2 are needed to get 36-2g of MnCl-2 from the equation given below- MnO2 nbsp-s- nbsp-4HClaq nbsp-MnCl2 nbsp-aq-2H2O nbsp-l-Cl2g87 nbsp-g nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-36-5 nbsp-g nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-126 nbsp-g nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-18 nbsp-g nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-71 nbsp-g nbsp-25

The correct option is A 25In the given equation 87g of nbsp;MnO_2 gives 126g of MnCl_2. To produce 1g of MnCl_2 the amount of nbsp;MnO_2 required = 87/126g ...

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Byju's Answer

Standard IX

Chemistry

Mass, Volume and Mole Relationship

How many gram...

Question

How many grams of ${MnO}_{2}$ are needed to get 36.2g of ${MnCl}_{2}$ from the equation given below?
${MnO}_{2} (s) + 4 HCl (aq) \to {MnCl}_{2} (aq) + 2 H_{2} O (l) + {Cl}_{2} (g) 87 g 36.5 g 126 g 18 g 71 g$
25

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Solution

The correct option is A 25
In the given equation 87g of ${MnO}_{2}$ gives 126g of ${MnCl}_{2}$ .

To produce 1g of ${MnCl}_{2}$ the amount of ${MnO}_{2}$ required = $\frac{87}{126}$ g

Hence, to produce 36.2 g of ${MnCl}_{2}$ the amount of ${MnO}_{2}$ required = $\frac{87}{126} \times 36.2$ = 25 (approx.)

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Similar questions

Q. How many grams of

{MnO}_{2}

are needed to get 36.2g of

{MnCl}_{2}

from the equation given below?

{MnO}_{2} (s) + 4 HCl (aq) \to {MnCl}_{2} (aq) + 2 H_{2} O (l) + {Cl}_{2} (g) 87 g 36.5 g 126 g 18 g 71 g

Q. How many grams of

{MnO}_{2}

are needed to get 36.2g of

{MnCl}_{2}

from the equation given below?

{MnO}_{2} (s) + 4 HCl (aq) \to {MnCl}_{2} (aq) + 2 H_{2} O (l) + {Cl}_{2} (g) 87 g 36.5 g 126 g 18 g 71 g

4HCl(aq) + MnO2(s) - 2H2O(L) + MnCl2(aq) + Cl2(g).

How many grams of HCl would react with 5 grams on MnO2, if Mn=55u and Cl=35.5u ?

Q. How many grams of HCl is needed for complete reaction with

69.6

M n O_{2}

H C l + M n O_{2} \to M n C l_{2} + H_{2} O + C l_{2}

Q. 10 g MnO2 on reaction with HCl forms 2.24 L of Cl2[g] at NTP, the percentage impurity of MnO2 is MnO2 + 4HCl - MnCl2 + Cl2 + 2H2O

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How many grams of MnO2 are needed to get 36.2g of MnCl2 from the equation given below? MnO2 (s)+ 4HCl(aq) →MnCl2 (aq)+2H2O (l)+Cl2(g)87 g 36.5 g 126 g 18 g 71 g 25

The correct option is A 25In the given equation 87g of MnO2 gives 126g of MnCl2. To produce 1g of MnCl2 the amount of MnO2 required = 87126g Hence, to produce 36.2 g of MnCl2 the amount of MnO2 required = 87126×36.2 = 25 (approx.)

How many grams of ${MnO}_{2}$ are needed to get 36.2g of ${MnCl}_{2}$ from the equation given below?
${MnO}_{2} (s) + 4 HCl (aq) \to {MnCl}_{2} (aq) + 2 H_{2} O (l) + {Cl}_{2} (g) 87 g 36.5 g 126 g 18 g 71 g$
25