Question

Stoichiometry! :) Please note:
- Use 6.022x1023 for avogadro’s number
- Ignore sig figs and do not round the final answer.
- Keep it to 1 decimal place.

Answer 1

Answer:

18.5 grams

Explanation:

The first step is to balance the equation.

2LiOH+CO2⇒Li2CO3+H2O+ some extra oxygen

For CO2, carbon has a molar mass of about 12, and oxygen has a molar mass of about 16, so the total is 44. Now that the equation is balanced and there are two LiOH's, it's molar mass is now doubled and it effectively has a molar mass of 48. 12/48=0.25 moles of both original substances used, and 0.25 moles of the final products. Li2CO3 has a molar mass of 7*2+12+16*3=74, and multiplying this by the 0.25 moles yields 18.5 grams. Hope this helps!

Answer 2

Answer:

18.5 grams $Li_{2}CO_{3}$

Explanation:

First, we'll need to balance the equation:

$LiOH_{s} + CO_{2(g)}$ → $Li_{2}CO_{3(s)} + H_{2}O_{(l)}$

There is one (Li) on the left and two on the right, so let's add a 2 coefficient on the left.

$2LiOH_{s} + CO_{2(g)}$ → $Li_{2}CO_{3(s)} + H_{2}O_{(l)}$

Now there are 2 (Li) on the left, and 2 on the right, 4 (O) on both sides, 1 (C) on both sides, and 2 (H) on both sides. The equation is balanced!

Step 1: Find the limiting reactant

To find the limiting reactant, we need to convert the given masses of each reactant into moles. Multiply the given masses by the molar masses.

12g 2LiOh × $\frac{1 mol}{48 g}$(double the molar mass because you have two molecules) = 0.25 moles

12g CO₂ × $\frac{1 mol}{44.01 g}$ = 0.272 moles

We'll test both reactants to see which one limits us.

• Given 0.25 mole LiOH × $\frac{1 mole CO_{2}}{2 moles LiOH}$ = .125 moles CO₂ needed
• Given 0.272 moles CO₂ x $\frac{2 moles LiOH}{1 mole CO_{2}}$ = 0.545 moles LiOH needed

Since we don't have enough CO₂ to use all of our LiOH, CO₂ is the limiting reactant. We will use all of the CO₂ to perform our reaction and ignore the excess LiOH.

Step 2: Calculating the mass of the product

You can find the mass of either product with a mole ratio. (Remember, we have too much CO₂, so we'll need to use the given LiOH to perform this calculation)

0.25 moles LiOH × $\frac{1 mole CO_{2}}{1 mole Li_{2}CO_{3}}$ = 0.25 moles $Li_{2}CO_{3}$

Now we convert back to grams!

0.25 moles $Li_{2}CO_{3}$ × $\frac{74 g}{1 mole}$ = 18.5 grams