A 6.11-g sample of a copper-aluminum alloy reacts with hydrochloric acid to produce 1.26 L of
hydrogen gas at 22 °C and 728 torr. Find the mass percentage of aluminum in the alloy sample. (An alloy is a material formed by mixing two or more metals. The alloy you are considering here contains only aluminum and copper.)
Ok so i used pv=nrt to find that there was .0498 mol of H2(hydrogen gas) prodcued. After this im stuck because i don't have a balanced equation to go off of. Is there a way to do it without an equation or is that the only way? The answer will need to be 14.7%
hydrogen gas at 22 °C and 728 torr. Find the mass percentage of aluminum in the alloy sample. (An alloy is a material formed by mixing two or more metals. The alloy you are considering here contains only aluminum and copper.)
Ok so i used pv=nrt to find that there was .0498 mol of H2(hydrogen gas) prodcued. After this im stuck because i don't have a balanced equation to go off of. Is there a way to do it without an equation or is that the only way? The answer will need to be 14.7%
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2Cu-Al + 6HCl --> 2Cu + 3H2 + 2AlCl3
Cu will NOT react with HCl as Cu is less reactive than H2
pv=nrt solving for n
n = pv/(rt)
n = 0.958atm x 1.26L / (0.0821L-atm/mole-K x 295K) = 0.05moles H2 produced
0.05moles H2 (2Al / 3H2) x 27g/mole = 0.897g Al
0.897g / 6.11g x 100% = 14.7%
Cu will NOT react with HCl as Cu is less reactive than H2
pv=nrt solving for n
n = pv/(rt)
n = 0.958atm x 1.26L / (0.0821L-atm/mole-K x 295K) = 0.05moles H2 produced
0.05moles H2 (2Al / 3H2) x 27g/mole = 0.897g Al
0.897g / 6.11g x 100% = 14.7%