Mg metal reacts with HCl to produce hydrogen gas. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) How many grams of magnesium are needed to prepare 8.00 L of H2 at 765 mmHg and 27 °C?
Added by Ronald R.
Step 1
We can use the ideal gas law equation, PV = nRT, to convert the given volume of H2 gas to moles. PV = nRT n = PV / RT n = (765 mmHg) * (8.00 L) / (0.0821 L·atm/mol·K * 300 K) n = 25.2 mol Show more…
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Mg metal reacts with HCl to produce hydrogen gas. $$ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{MgCl}_{2}(a q) $$ a. What volume, in liters, of hydrogen at $0{ }^{\circ} \mathrm{C}$ and $1.00 \mathrm{~atm}$ (STP) is released when $8.25 \mathrm{~g}$ of $\mathrm{Mg}$ reacts? b. How many grams of magnesium are needed to prepare 5.00 L of $\mathrm{H}_{2}$ at $735 \mathrm{mmHg}$ and $18^{\circ} \mathrm{C}$ ?
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