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How many mL of 6.00 M HCl(aq) solution are required to completely consume a 37.5 g sample of zinc metal (atomic weight = 65.39 g/mol) if the reaction is Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)?

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)?

Solution: We need to use molar mass of Zn (given in the question) to find moles of Zn and then use the coefficients in the equation to go from moles of Zn to moles of HCl. There are 2 moles of HCl for 1 mol of Zn according to the balanced equation given.

37.5g Zn( 1molZn/65.39gZn)( 2mol HCl/1mol Zn) = 1.14696mol HCl

Molarity = mol/L

We have molarity (given in the question) and we calculated the moles. We can plug both in to find Liters of HCl and then convert liters to milliliters.

6.00M HCl = 1.14696mol HCl/x

x = 0.19116 L

0.19116 L (1000ml/1L) = 191ml

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