Question

we wish to determine the mass of baso₄ formed when 200. ml of 0.10 m bacl₂ reacts with excess na₂so₄ according to the equation below.
bacl₂(aq) + na₂so₄(aq) → baso₄(s) + 2nacl(aq)
how many moles of bacl₂ are present
in 200. ml of 0.10 m bacl₂?

Explanation:

Step1: Recall Molarity Formula

Molarity ($M$) is defined as moles of solute ($n$) per liter of solution ($V$ in liters), so the formula is $M=\frac{n}{V}$. We need to find $n$, so rearrange to $n = M\times V$.

Step2: Convert Volume to Liters

The volume is 200. mL. Since 1 L = 1000 mL, convert 200. mL to liters: $V=\frac{200.}{1000}=0.200\space L$.

Step3: Calculate Moles of $\boldsymbol{BaCl_2}$

Given $M = 0.10\space M$ (or $0.10\space mol/L$) and $V = 0.200\space L$, use $n = M\times V$. So $n=0.10\space mol/L\times0.200\space L = 0.020\space mol$.

Answer:

The number of moles of $\ce{BaCl_{2}}$ is $\boldsymbol{0.020\space mol}$ (or 0.02 mol, depending on significant figures; 200. mL has three, 0.10 M has two, so two significant figures: 0.020 mol or 0.02 mol? Wait, 200. mL is three, 0.10 is two, so the result should have two. Wait, 0.10 0.200 = 0.020 (three? No, 0.10 is two, 0.200 is three, so the least is two? Wait, molarity: 0.10 M is two significant figures, volume 200. mL is three (the decimal after 200 means it's three). So when multiplying, the number of significant figures is determined by the least, which is two. But 0.10 0.200 = 0.020 (exactly, because 0.10 is 1.0 x 10^-1, 0.200 is 2.00 x 10^-1, so 1.02.00 = 2.00 x 10^-2, which is 0.020 mol (two significant figures? Wait, 0.10 has two, 0.200 has three, so the product should have two. Wait, 0.10 0.200 = 0.020 (the trailing zero after 2 is significant because 0.10 has a zero after the decimal, so 0.020 is two significant figures? Wait, no: 0.020 has two significant figures (the 2 and the trailing zero). So the correct calculation gives 0.020 moles (or 2.0 x 10^-2 mol).

Wait, let's do it again: $n = M \times V$. $M = 0.10\space mol/L$, $V = 200.\space mL = 0.200\space L$. So $n = 0.10\space mol/L \times 0.200\space L = 0.020\space mol$. Yes, because 0.10 0.200 = 0.020 (the zero after the 2 is significant here because the molarity has two significant figures and the volume has three, but when multiplying, the result should have two? Wait, no: 0.10 is two sig figs, 0.200 is three. The rule is that the number of sig figs in the result is equal to the least number of sig figs in the inputs. So two sig figs. But 0.10 0.200 = 0.020, which is two sig figs (the 2 and the trailing zero). So the answer is 0.020 moles (or 0.02 mol, but 0.020 is more accurate with the sig figs from the volume's decimal).

So the final answer is $\boldsymbol{0.020\space mol}$ (or 0.02 mol, but 0.020 is better here).