Question

how many molecules are in 4.50 mol of na₂co₃?
7.21*10²⁴
271
2.71*10²⁴
2.00*10²⁴
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20 multiple choice 1 point
in a laboratory, the following reaction was conducted: 5 mol of h₂ reacted with o₂ in excess, and 25 grams of water h₂o were recovered. what is the actual yield for this reaction?
25 grams
25 mol
5 mol
5 grams
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Explanation:

First Question (Molecules in 4.50 mol of $\boldsymbol{\ce{Na2CO3}}$)

Step1: Recall Avogadro's Number

Avogadro's number is $N_A = 6.022\times 10^{23}\ \text{molecules/mol}$. To find the number of molecules, use the formula $N = n\times N_A$, where $n$ is the number of moles.

Step2: Substitute Values

Given $n = 4.50\ \text{mol}$, so $N = 4.50\ \text{mol} \times 6.022\times 10^{23}\ \text{molecules/mol}$.
Calculate $4.50\times 6.022\times 10^{23} = 27.099\times 10^{23} = 2.71\times 10^{24}$ (after rounding to three significant figures).

Actual yield is the amount of product actually obtained from a reaction. The problem states 25 grams of $\ce{H2O}$ were recovered, so the actual yield is 25 grams.

Answer:

$\boldsymbol{2.71\times 10^{24}}$ (corresponding to the option "2.71*10²⁴")

Second Question (Actual Yield of $\boldsymbol{\ce{H2O}}$)