Question

For each reaction, calculate the mass of the product that forms when 14.4 g of the reactant in red completely reacts. Assume that there is more than enough of the other reactant.

2K(s)+Cl2(g)→2KCl(s)
Express your answer in grams to three significant figures.

2K(s)+Br2(l)→2KBr(s)
Express your answer in grams to three significant figures.

4Cr(s)+3O2(g)→2Cr2O3(s)
Express your answer in grams to three significant figures.

2Sr(s)+O2(g)→2SrO(s)
Express your answer in grams to three significant figures.

Answer 1

Answer on question #62247, Chemistry / General Chemistry

For each reaction, calculate the mass of the product that forms when $14.4\mathrm{g}$ of the reactant in red completely reacts. Assume that there is more than enough of the other reactant.

$$2 \mathrm {K} (\mathrm {s}) + \mathrm {Cl} _ {2} (\mathrm {g}) \rightarrow 2 \mathrm {KCl} (\mathrm {s})$$

Express your answer in grams to three significant figures.

$$2 \mathrm {K} (\mathrm {s}) + \mathrm {Br} _ {2} (\mathrm {l}) \rightarrow 2 \mathrm {KBr} (\mathrm {s})$$

Express your answer in grams to three significant figures.

$$4 \mathrm {Cr} (\mathrm {s}) + 3 \mathrm {O} _ {2} (\mathrm {g}) \rightarrow 2 \mathrm {Cr} _ {2} \mathrm {O} _ {3} (\mathrm {s})$$

Express your answer in grams to three significant figures.

$$2 \mathrm{Sr}(\mathrm {s}) + \mathrm{O} _ {2} (\mathrm {g}) \rightarrow 2 \mathrm{SrO}(\mathrm {s})$$

Express your answer in grams to three significant figures.

Solution:

$$2 K (s) + C l _ {2} (g) \rightarrow 2 K C l (s)$$

moles $\mathrm{Cl}_2 = 14.4\mathrm{g} / 70.906\mathrm{g / mol} = 0.203\mathrm{mol}$

moles KCl produced $= 2 \times 0.203$ mol=0.406 mol

mass $\mathrm{KCl} = 0.406\mathrm{mol}\times 78.196\mathrm{g / mol} = 31.7\mathrm{g}$

Answer: 31.7 g

$$2 K (s) + B r _ {2} (l) \rightarrow 2 K B r (s)$$

moles $\mathrm{Br}_2 = 14.4\mathrm{g} / 159.808\mathrm{g / mol} = 0.0901$

moles KBr = 2 x 0.0901 =0.180

mass $\mathrm{KBr} = 0.180\mathrm{mol}\times 119.0\mathrm{g / mol} = 21.4\mathrm{g}$

Answer: 21.4 g

$$4 C r (s) + 3 O _ {2} (g) \rightarrow 2 C r _ {2} O _ {3} (s)$$

moles $\mathrm{O}_2 = 14.4\mathrm{g} / 32\mathrm{g / mol} = 0.450$

moles $\mathrm{Cr_2O_3} = 0.450\times 2 / 3 = 0.300$

mass $\mathrm{Cr_2O_3} = 0.300\mathrm{mol}\times 151.99\mathrm{g / mol} = 45.6\mathrm{g}$

Answer: 45.6 g

$$2 S r (s) + O _ {2} (g) \rightarrow 2 S r O (s)$$

moles $\mathrm{Sr} = 14.4\mathrm{g} / 175.24\mathrm{g / mol} = 0.0822$

moles $\mathrm{SrO} = 2\times 0.0822 = 0.1644$

mass $\mathrm{SrO} = 0.1644\mathrm{mol}\times 103.62\mathrm{g / mol} = 17.0\mathrm{g}$

Answer: 17.0 g

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