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2 years ago

Given the reaction below, if 0.00345 g of carbon dioxide is used up, how many grams of water will be produced?

Chemistry

1 answer:

zimovet [89]2 years ago

3 0

From the stoichiometry of the reaction, 1.4 * 10^-3 g is produced.

<h3>What mass of water is produced?</h3>

The equation of the reaction is written as; CO2 + 2LiOH → Li2CO3 + H2O. This can help us to apply the principle of stoichiometry here.

Thus;

Number of moles of CO2 = 0.00345 g/44 g/mol = 7.8 * 10^-5 moles

If 1 mole of CO2 produced 1 mole of water

7.8 * 10^-5 moles of CO2 produced 7.8 * 10^-5 moles of water

Mass of water produced = 7.8 * 10^-5 moles * 18 g/mol = 1.4 * 10^-3 g

Learn ore about stoichiometry:brainly.com/question/9743981

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${ \sf{1 \: l \: of \: KOH \: contains \: 0.3021 \: moles}} \\ { \sf{0.02894 \: l \: of \: KOH \: contain \: (0.02894 \times 0.3021) \: moles}} \\ { \underline{ = 0.008743 \: moles}}$

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.

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${ \sf{0.008743 \: moles \: of \: acid \: weigh \: 3.7066 \: g}} \\ { \sf{1 \: mole \: of \: acid \: weighs \: ( \frac{1 \times 3.7066}{0.008743}) \: g }} \\ = { \underline{423.95 \: g \approx424 \: grams}}$

for the molar mass:

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