What volume of hydrogen gas (measured at STP) would result from reacting 75.0g of sodium hydroxide with 50.0g of aluminum? 6NaOH + 2Al --> 2Na3AlO3 + 3H2

#6"NaOH" + 2"Al" -> 2"Na"_3"AlO"_3 + 3"H"_2#

1. First step would be to balance the chemical equation, which is already done for us

2. Determine the limiting reagent using stoichometric ratios
   #75.0 cancel"g NaOH"xx (1 cancel"mol NaOH")/(40.0 cancel"g NaOH")xx(2 cancel"mol Al")/(6 cancel"mol NaOH")xx(26.98 "g Al")/ (1cancel"mol Al")#= 16.9 g Al needed to react with this amount of NaOH

And since we have 50 g of Al in the beginning, Al is in excess and NaOH is limiting
Note: Molar Volume at STP is about #22.4 L/"mol"#
3. Use the limiting reagent to determine product output using stoichometric ratios
#75.0 cancel"g NaOH"xx (1 cancel"mol NaOH")/(40.0 cancel"g NaOH")xx(3 cancel("mol" H_"2"))/(6 cancel"mol NaOH")xx(22.4L H_2)/(cancel (mol H_2))= 21.0" L H"_2#

Balanced equation

#"6NaOH + 2Al"##rarr##"2Na"_3"AlO"_3 + "3H"_2"#

First you need to find the limiting reactant. The reactant that produces the least amount of hydrogen is the limiting reactant.

Then use the molar volume of a gas at STP to determine the volume of #"H"_2"# that can be produced.

To find the limiting reactant, determine mol #"H"_2"# that can be produced by each reactant.

Determine mol reactant by dividing the given mass by its molar mass. Do this by multiplying the given mass of the reactant by the inverse of its molar mass. To determine mol #"H"_2"#, multiply by the mol ratio between the reactant and #"H"_2"# from the balanced equation, with #"H"_2"# in the numerator.

#75.0color(red)cancel(color(black)("g NaOH"))xx(1color(red)cancel(color(black)("mol NaOH")))/(39.997color(red)cancel(color(black)("g NaOH")))xx(3"mol H"_2)/(6color(red)cancel(color(black)("mol NaOH")))="0.938 H"_2"#

#50.0color(red)cancel(color(black)("g Al"))xx(1color(red)cancel(color(black)("mol Al")))/(26.982color(red)cancel(color(black)("g Al")))xx(3"mol H"_2)/(2color(red)cancel(color(black)("mol Al")))="2.78 mol H"_2"#

Sodium hydroxide produces the least amount of hydrogen gas, so it is the limiting reactant.

Current #"STP"# is #0^@"C"# or #"273.15 K"#, and #10^5# #"Pa"# or #"100 kPa"#.

The molar volume of a gas at STP is #"22.711 L/mol"#.

Multiply mol #"H"_2"# by the molar volume.

#0.938color(red)cancel(color(black)("mol H"_2))xx(22.711"L")/(1color(red)cancel(color(black)("mol")))="21.3 L H"_2"#

If your instructor wants you to use #"1 atm"# for standard pressure, the molar volume at STP is #"22.414 L/mol"#.