If 26.5 mL of 0.222 M HCl reacts with excess Mg, how many mL of hydrogen gas are produced at STP?
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
4 answers
duplicate post.
26.5 *10^-3 L * 0.222 mol / L H+ = 8.1*10^-3 mol H+ or 4.05*10^-3 mol H2
4.05*10^-3 mol H2 * 22.4 Liters/mol at STP = 90.7 *10^-3 liters
= 90.7 mL
4.05*10^-3 mol H2 * 22.4 Liters/mol at STP = 90.7 *10^-3 liters
= 90.7 mL
calc error
26.5 *10^-3 L * 0.222 mol / L H+ = 5.88*10^-3 mol H+ or 2.94*10^-3 mol H2
2.94*10^-3 mol H2 * 22.4 Liters/mol at STP = 65.9 *10^-3 liters
= 65.9 mL
26.5 *10^-3 L * 0.222 mol / L H+ = 5.88*10^-3 mol H+ or 2.94*10^-3 mol H2
2.94*10^-3 mol H2 * 22.4 Liters/mol at STP = 65.9 *10^-3 liters
= 65.9 mL
Nope. I suspect you punched in the wrong numbers since 26.5E-3 x 0.222 = 5.88*10^-3 and half that is 2.94*10^-3 mols H2 gas @ STP
1 answer