19.5k views
5 votes
What mass in grams of tin would be required to completely react with 1.45 L of 0.750 M HBr in the following chemical reaction?

Sn(s) + 4 HBr(aq) -› SnBr, (aq) + 2 H, (g)

User Thusitha
by
8.3k points

1 Answer

4 votes
The balanced chemical equation for the reaction is:

Sn(s) + 4 HBr(aq) → SnBr4(aq) + 2 H2(g)

From the equation, we can see that 1 mole of Sn reacts with 4 moles of HBr. We can use this information to calculate the number of moles of HBr in 1.45 L of 0.750 M solution:

0.750 M = 0.750 moles/L
1.45 L x 0.750 moles/L = 1.0875 moles HBr

According to the stoichiometry of the reaction, 1 mole of Sn reacts with 4 moles of HBr. Therefore, the number of moles of Sn required to react with 1.0875 moles of HBr is:

1.0875 moles HBr x (1 mole Sn / 4 moles HBr) = 0.2719 moles Sn

Finally, we can calculate the mass of Sn required using its molar mass:

0.2719 moles Sn x 118.71 g/mol = 32.3 g Sn

Therefore, 32.3 grams of tin would be required to completely react with 1.45 L of 0.750 M HBr in the given chemical reaction.

Hoped this helped!
User Cgons
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.