5.0g = 5.0/26.98 = 0.185 moles Al
50 mL of 6M = 0.3 moles HCl
Since 2Al + 6HCl --> 2AlCl3 + 3H2, only 0.1 moles of Al is used, producing 0.1 moles AlCl3 and 0.15 moles H2.
so, the volume of gas is 0.15 * 22.4L
for the other problem, use PV=nRT to find the # moles. Then divide into the mass.
5.0 g of aluminum reacts with 50.0 mL of 6.0 M hydrochloric acid to produce
hydrogen gas. What is the volume of gas collected if the gas was collected at STP?
3. A 0.418 g sample of gas has a volume of 115 mL at 66.3 °C and 743 mmHg. What is
the molar mass of this gas?
1 answer
1 answer