Question

Aluminum metal reacts with manganese(II) chloride to form manganese metal and aluminum chloride. If 9.80 g of aluminum are reacted with 70.0 g of manganese(II) chloride how many g of the reagent in excess will remain at the end of the reaction? 1.40 g 4.80 g none of the answers are correct 4.20 g 0.500 g

          Aluminum metal reacts with manganese(II) chloride to form manganese metal and aluminum chloride. If 9.80 g of aluminum are reacted with 70.0 g of manganese(II) chloride how many g of the reagent in excess will remain at the end of the reaction?
1.40 g
4.80 g
none of the answers are correct
4.20 g
0.500 g

Aluminum metal reacts with manganese(II) chloride to form manganese metal and aluminum chloride. If 9.80 g of aluminum are reacted with 70.0 g of manganese(II) chloride how many g of the reagent in excess will remain at the end of the reaction?
1.40 g
4.80 g
none of the answers are correct
4.20 g
0.500 g

Added by Jessica T.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Ma Ednelyn Lim

03/30/2022

Step 1

Given: Mass of aluminum = 9.80 g Molar mass of aluminum = 26.98 g/mol Moles of aluminum = 9.80 g / 26.98 g/mol = 0.363 mol Show more…

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Aluminum metal reacts with manganese(ll) chloride to form manganese metal and aluminum chloride. If 9.80 g of aluminum are reacted with 70.0 g of manganese(ll) chloride how many g of the reagent in excess will remain at the end of the reaction? 1.40 g 4.80 g none of the answers are correct 4.20 g 0.500 g