A 5.00 g sample of Al reacts with excess H_2SO_4. What volume of H_2 gas is prod

Question

A 5.00 g sample of Al reacts with excess H_2SO_4. What volume of H_2 gas is produced at STP? 2 Al(s) + 3 H_2SO_4(aq) rightarrow Al_2(SO_4)3 (aq) + 3 H_2(g) Two solutions. 250.0 mL of 1.00 M CaCl_2(aq) and 250.0 mL of 1.00 M K_2CO_3(aq), are added to a calorimeter with a heat capacity of 1.25 kJ/degree C and the temperature decreased by 2.40 degree C. Determine Delta H degree_rxn per mole of CaCO_3(s) formed in the reaction. CaCl_2(aq) + K_2CO_3(aq) rightarrow CaCO_3(s) + 2 KCl (aq) For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac).

Explanation / Answer

A 5. 00 g sample of Al REACTS with excess H2SO4 . WHAT VOLUME OF H2 GAS IS PRODUCED

2Al + 3H2SO4 ==> 3H2 + Al2(SO4)

2 MOLES OF Al REACTS WITH 3 MOLES OF H2SO4 GIVES 3 MOLES OF H2 + Al2(SO4)

HENCE

NUMBER OF MOLES OF Aluminium = 5.00 g Al x (1 mole Al / 27.0 g Al) = 0.185 moles Aluminium

NUMBER OF MOLES H2 PRODUCED==0.185 moles Al x (3 moles H2 / 2 moles Al) = 0.278 moles H2

AT STP

VOLUME OF ONE MOLE OF A GAS ---------- 22.4 L.
VOLUME OF H2 PRODUCED

= 0.278 moles H2 x (22.4 L H2 / 1 mole H2) = 6.23 L H2

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