Question

In the reaction Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq), how many moles of hydrogen gas will be produced from 250.0 milliliters of a 3.0 M HCl in an excess of Mg? 0.75 moles 0.38 moles 3.0 moles 1.5 moles

Answer 1

Mg + 2HCl = H₂ + MgCl₂

n(HCl)=c(HCl)v(HCl)

n(H₂)=n(HCl)/2=c(HCl)v(HCl)/2

n(H₂)=3.0mol/L*0.2500L/2=0.375 mol≈0.38 mol

Answer 2

Answer:

0.38 moles of hydrogen gas will be produced.

Explanation:

$Mg(s) + 2HCl(aq)\rightarrow H_2(g) + MgCl_2(aq)$

Mole sof HCl = n

Molarity of the HCL solution = 3.0 M

Volume of HCL solution = 250.0 mL = 0.250 L

$Molarity=\frac{\text{moles of solute}}{\text{Volume of the solution(L)}}$

$3.0 M=\frac{n}{0.250 L}$

n = 0.75 mol

According to reaction , 2 moles of HCl gives 1 mol of hydrogen gas.

Then 0.75 mol of HCL will give:

$\frac{1}{2}\times 0.75 mol=0.375 mol\approx 0.38 mol$ of hydrogen gas.

0.38 moles of hydrogen gas will be produced from 250.0 milliliters of a 3.0 M HCl in an excess of Mg.