What volume of hydrogen in ml will be liberated at STP by the reaction of Zn on 50 ml of dilute H2SO4 of specific gravity 1.3 having purity of 40%? Report your answer by rounding it upto nearest whole number.

Question

What volume of hydrogen in ml will be liberated at STP by the reaction of Zn on 50 ml of dilute H2SO4 of specific gravity 1.3 having purity of 40%? Report your answer by rounding it upto nearest whole number. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Mass of H2SO4=(50 × 13 × 40/100)=26 text g Zn+ underset underseto r text 1 mole98 text gH2 S O4 arrow ZnSO4-+ underset underseta t text NTP22.4 text LH2 As 98 gm H2SO4 provides =22.4 text L textH2 So 26 gm H2SO4 provides =(22.4 × 26/98) =5.95 text L ≈ 6 text L

Q. What volume of hydrogen in ml will be liberated at STP by the reaction of Zn on 50 ml of dilute $H_{2} S O_{4}$ of specific gravity 1.3 having purity of 40%?

Report your answer by rounding it upto nearest whole number.

Mass of $H_{2} S O_{4} = \frac{50 \times 13 \times 40}{100} = 26 g$

$Z n + or 1 mole 98 g H_{2} S O_{4} \to Z n S O_{4}^{-} + a t NTP 22.4 L H_{2}$

As 98 gm $H_{2} S O_{4}$ provides $= 22.4 L H_{2}$

So 26 gm $H_{2} S O_{4}$ provides $= \frac{22.4 \times 26}{98}$

$= 5.95 L$

$\approx 6 L$

Q. What volume of hydrogen in ml will be liberated at STP by the reaction of Zn on 50 ml of dilute H2​SO4​ of specific gravity 1.3 having purity of 40%? Report your answer by rounding it upto nearest whole number.

Mass of H2​SO4​=10050×13×40​=26 g Zn+or 1 mole98 g​H2​SO4​​→ZnSO4−​+at NTP22.4 L​H2​​ As 98 gm H2​SO4​ provides =22.4 L H2​ So 26 gm H2​SO4​ provides =9822.4×26​ =5.95 L ≈6 L

Q. What volume of hydrogen in ml will be liberated at STP by the reaction of Zn on 50 ml of dilute $H_{2} S O_{4}$ of specific gravity 1.3 having purity of 40%?

Report your answer by rounding it upto nearest whole number.

Mass of $H_{2} S O_{4} = \frac{50 \times 13 \times 40}{100} = 26 g$

$Z n + or 1 mole 98 g H_{2} S O_{4} \to Z n S O_{4}^{-} + a t NTP 22.4 L H_{2}$

As 98 gm $H_{2} S O_{4}$ provides $= 22.4 L H_{2}$

So 26 gm $H_{2} S O_{4}$ provides $= \frac{22.4 \times 26}{98}$

$= 5.95 L$

$\approx 6 L$