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What volume of hydrogen gas (measured at STP) would result from reacting 75.0g of sodium hydroxide with 50.0g of aluminum

User Tworec
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4 votes

Answer:

82.9 L of H2

Step-by-step explanation:

We need to start with a balanced equation. The following looks correct:

2NaOH+2Al+2H2O = 2NaAlO2 + 3H2

This tells us to expect 3 moles of H2 gas for every 2 moles of sodium hydroxide. It also tells us that we need 2 moles of Al for every 2 moles of NaOH.

Now calculate the moles of both the sodium hydroxide and the aluminum:

NaOH: The molar mass is 40.0 g/mole

(75g NaOH)/(40.0 g/mole) = 1.875 moles NaOH

Al: The molar mass is 27.0 g/mole

(50.0g Al)/(27.0 g/mole) = 1.85 moles Al

The values are close, but the reaction will stop when all the moles of Al are consumed, leaving 0.025 moles of NaOH, since they are consumed at the same proportion, as per the balanced equation.

The equation says we should expect 3 moles H2 gas for evey 2 moles of Al. So:

(1.85 moles Al)*(3 moles H2)/(2 moles Al) = 3.7 moles of H2

At STP, the conversion factor for all gasses is 22.4L/mole (STP only).

(3.7 moles of H2)*(22.4L/mole) = 82.9 L of H2

User Sam Krygsheld
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