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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
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1 Answer

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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
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