Chemistry, 05.05.2020 12:57 cranfordjacori

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.

Stoichiometry! :) Please note: - Use 6.022x1023 for avogadro’s number - Ignore sig figs and do not

Answers: 3

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Answer from: Marshmallow9174

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 from: kaseyvn03

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 neededGiven 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

Answer from: Quest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$

Answer from: Quest

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.

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Stoichiometry! :) Please note:
- Use 6.022x1023 for avogadro’s number
- Ignore sig figs...