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Question

# How many moles of excess reagent are left by the reaction of 6.4 mol of HCl and 3.5 mol of Zn? The reaction that takes place is: Zn+2HCl→ZnCl2+H2

A
0.30 mol
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B
3.45 mol
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C
4.90 mol
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D
0.54 mol
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Solution

## The correct option is A 0.30 molZn+2HCl→ZnCl2+H2 To find limiting reagent: For Zn=3.51=3.5 For HCl=6.42=3.2 From above, we can say that HCl is the limiting reagent and Zn is the excess reagent. To find the number of moles of Zn required to consume 6.4 mol HCl: 6.4 mol of HCl×1 mol of ZnCl22 mol of HCl=3.2 mol of Zn Number of moles of the excess reagent left = 3.5 - 3.2 = 0.3 mol

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