Question

period
structions: determine the number of grams in each of the quantities below. show your
2.5 moles of nacl

0.50 moles of h₂so₄

1.70 moles of kmno₄

0.25 moles of kcl

what is the mass of 5 moles of fe₂o₃ ?

find the grams in 1.26 x 10⁻⁴ mol of hc₂h₃o₂.

find the mass in 2.6 mol of lithium bromide.

what is the mass in grams of 1.56 moles of aspartame c₁₄h₁₈n₂o₅

2.0 moles of aucl₃ is equal to how many grams?

4.0 moles of h₂o would have a mass of ______?

how many grams are in 4.5 moles of naf ?

how many grams are in 0.2 moles of bei₂ ?

Explanation:

Step1: Calculate molar mass of NaCl

Molar mass of $\text{NaCl} = 23 + 35.5 = 58.5\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 2.5\ \text{mol} \times 58.5\ \text{g/mol} = 146.25\ \text{g}$

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Step1: Calculate molar mass of $\text{H}_2\text{SO}_4$

Molar mass of $\text{H}_2\text{SO}_4 = 2(1) + 32 + 4(16) = 98\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 0.50\ \text{mol} \times 98\ \text{g/mol} = 49\ \text{g}$

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Step1: Calculate molar mass of $\text{KMnO}_4$

Molar mass of $\text{KMnO}_4 = 39 + 55 + 4(16) = 158\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 1.70\ \text{mol} \times 158\ \text{g/mol} = 268.6\ \text{g}$

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Step1: Calculate molar mass of $\text{KCl}$

Molar mass of $\text{KCl} = 39 + 35.5 = 74.5\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 0.25\ \text{mol} \times 74.5\ \text{g/mol} = 18.625\ \text{g}$

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Step1: Calculate molar mass of $\text{Fe}_2\text{O}_3$

Molar mass of $\text{Fe}_2\text{O}_3 = 2(56) + 3(16) = 160\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 5\ \text{mol} \times 160\ \text{g/mol} = 800\ \text{g}$

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Step1: Calculate molar mass of $\text{HC}_2\text{H}_3\text{O}_2$

Molar mass of $\text{HC}_2\text{H}_3\text{O}_2 = 2(12) + 4(1) + 2(16) = 60\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 1.26 \times 10^{-4}\ \text{mol} \times 60\ \text{g/mol} = 7.56 \times 10^{-3}\ \text{g}$

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Step1: Calculate molar mass of $\text{LiBr}$

Molar mass of $\text{LiBr} = 7 + 79.9 = 86.9\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 2.6\ \text{mol} \times 86.9\ \text{g/mol} = 225.94\ \text{g}$

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Step1: Calculate molar mass of $\text{C}_{14}\text{H}_{18}\text{N}_2\text{O}_5$

Molar mass of $\text{C}_{14}\text{H}_{18}\text{N}_2\text{O}_5 = 14(12) + 18(1) + 2(14) + 5(16) = 294\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 1.56\ \text{mol} \times 294\ \text{g/mol} = 458.64\ \text{g}$

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Step1: Calculate molar mass of $\text{AuCl}_3$

Molar mass of $\text{AuCl}_3 = 197 + 3(35.5) = 303.5\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 2.0\ \text{mol} \times 303.5\ \text{g/mol} = 607\ \text{g}$

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Step1: Calculate molar mass of $\text{H}_2\text{O}$

Molar mass of $\text{H}_2\text{O} = 2(1) + 16 = 18\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 4.0\ \text{mol} \times 18\ \text{g/mol} = 72\ \text{g}$

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Step1: Calculate molar mass of $\text{NaF}$

Molar mass of $\text{NaF} = 23 + 19 = 42\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 4.5\ \text{mol} \times 42\ \text{g/mol} = 189\ \text{g}$

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Step1: Calculate molar mass of $\text{BeI}_2$

Molar mass of $\text{BeI}_2 = 9 + 2(126.9) = 262.8\ \text{g/mol}$

Step2: Convert moles to grams

$\text{Mass} = 0.2\ \text{mol} \times 262.8\ \text{g/mol} = 52.56\ \text{g}$

Answer:

1. 2.5 moles of NaCl: $\boldsymbol{146.25\ \text{g}}$
2. 0.50 moles of $\text{H}_2\text{SO}_4$: $\boldsymbol{49\ \text{g}}$
3. 1.70 moles of $\text{KMnO}_4$: $\boldsymbol{268.6\ \text{g}}$
4. 0.25 moles of KCl: $\boldsymbol{18.625\ \text{g}}$
5. 5 moles of $\text{Fe}_2\text{O}_3$: $\boldsymbol{800\ \text{g}}$
6. $1.26 \times 10^{-4}$ mol of $\text{HC}_2\text{H}_3\text{O}_2$: $\boldsymbol{7.56 \times 10^{-3}\ \text{g}}$
7. 2.6 mol of lithium bromide: $\boldsymbol{225.94\ \text{g}}$
8. 1.56 moles of aspartame: $\boldsymbol{458.64\ \text{g}}$
9. 2.0 moles of $\text{AuCl}_3$: $\boldsymbol{607\ \text{g}}$
10. 4.0 moles of $\text{H}_2\text{O}$: $\boldsymbol{72\ \text{g}}$
11. 4.5 moles of NaF: $\boldsymbol{189\ \text{g}}$
12. 0.2 moles of $\text{BeI}_2$: $\boldsymbol{52.56\ \text{g}}$