A 3.50 g sample of solid Mg(OH) 2 was treated with50.0 mL of 0.500 M H 2 SO 4(aq

Question

A 3.50 g sample of solid Mg(OH)2 was treated with50.0 mL of 0.500 M H2SO4(aq). After thereaction was over, some of Mg(OH)2 remainedundissolved. Calculate how many g of solid Mg(OH)2remained. (Hint: start with a balanced equation). A) 1.00 g B) 2.04 g C) 1.46 g D) 3.47 g E) 1.75 g Please explain how you came to the answer. Any helpwould be appreciated! A 3.50 g sample of solid Mg(OH)2 was treated with50.0 mL of 0.500 M H2SO4(aq). After thereaction was over, some of Mg(OH)2 remainedundissolved. Calculate how many g of solid Mg(OH)2remained. (Hint: start with a balanced equation). A) 1.00 g B) 2.04 g C) 1.46 g D) 3.47 g E) 1.75 g Please explain how you came to the answer. Any helpwould be appreciated!

Explanation / Answer

1) Write the balanced equation, in this question it reallydoesn't matter - Mg(OH)2 + H2SO4--> MgSO4 + 2H2O
2) convert 3.5 g Mg(OH)2 to moles which is 0.06
3) Convert 50 mL/ 0.5M to moles, which is 0.025
4) Determine limiting reactant, which is H2SO4 in thiscase
5) So, if we only used 0.025 moles of H2SO4, that means weonly used 0.025 moles of Mg(OH)2
6) What's left? we substract the moles of Mg(OH)2 by moles ofH2SO4, so 0.06-0.025= 0.035 moles Mg(OH)2 left
7) convert 0.035 moles to grams which gives 2.04 g

hope this helps

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