What is the minimum amount of 6.0 M H2S04 necessary to produce 25.0 g of H2 (g)

Question

What is the minimum amount of 6.0 M H2S04 necessary to produce 25.0 g of H2 (g) according to the reaction between alurqinum and sulfuric acid?
2 Al(s-) + 3 H2S04(aq)-> Al2(S04)3(aq) + 3 H2(g) What is the minimum amount of 6.0 M H2S04 necessary to produce 25.0 g of H2 (g) according to the reaction between alurqinum and sulfuric acid?
2 Al(s-) + 3 H2S04(aq)-> Al2(S04)3(aq) + 3 H2(g) What is the minimum amount of 6.0 M H2S04 necessary to produce 25.0 g of H2 (g) according to the reaction between alurqinum and sulfuric acid?
2 Al(s-) + 3 H2S04(aq)-> Al2(S04)3(aq) + 3 H2(g)

Explanation / Answer

2 Al(s-) + 3 H2S04(aq)-> Al2(S04)3(aq) + 3 H2(g)

moles of H2 =grams /molar mass H2 = 25.0 g /2.0 = 12.5 moles

moles of H2SO4 required = ( 3mole H2SO4 / 3 mole H2) * 12.5 moles   = 12.5 moles

grams of H2SO4 = moles H2SO4 * molar mass H2SO4

grams H2SO4 = 12.5 * 98.0

= 1225 g H2SO4 (answer)

we need 1225 g H2SO4

in case you need volume of H2SO4

then we are given molarity as 6M

volume (in litres) = moles of H2SO4 / molarity = 12.5 moles / 6

= 2.08 L (answer)

In erms of volume ,we need 2.08 L H2SO4 (answer)

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