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Chapter 9: Problem 74

Consider the balanced chemical equation \(\mathrm{SCl}_{4}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{SO}_{2}+4 \mathrm{HCl}\) (a) How many grams of \(\mathrm{H}_{2} \mathrm{O}\) will react with \(5.000 \mathrm{~g}\) of \(\mathrm{SCl}_{4} ?\) (b) How many grams of \(\mathrm{SO}_{2}\) can you make from \(10.00 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} ?\) (c) Suppose you react \(5.000 \mathrm{~g}\) of \(\mathrm{SCl}_{4}\) with the amount of \(\mathrm{H}_{2} \mathrm{O}\) you calculated in part (a). How many grams of \(\mathrm{HCl}\) will you form?

Short Answer

Expert verified
(a) 1.0116 g of H2O will react with 5.000 g of SCl4. (b) 17.81 g of SO2 can be made from 10.00 g of H2O. (c) 4.1026 g of HCl will be formed when 5 g of SCl4 react with 1.0116 g of H2O.

Step by step solution

01

Convert given mass into moles

First, calculate the molar mass of the substances involved. For SCl4: \( M_{SCl4} = 1*32.1 + 4*35.5 = 178.1 g/mol \), for H2O: \( M_{H2O} = 2*1.0 + 1*16.0 = 18.0 g/mol\), and for SO2: \( M_{SO2} = 1*32.1 + 2*16.0 = 64.1 g/mol \). Next, convert the mass of SCl4 and H2O into moles. For SCl4: \(n_{SCl4} = \frac{5.00\mathrm{~g}}{178.1\mathrm{~g/mol}} = 0.0281\mathrm{~mol}\) For H2O: \(n_{H2O} = \frac{10.00\mathrm{~g}}{18.0\mathrm{~g/mol}} = 0.5556\mathrm{~mol}\)
02

Calculate moles of H2O needed for SCl4 reaction (part a)

From the balanced chemical equation, we can see that 1 mole of SCl4 requires 2 moles of H2O. So, to find the moles of water needed for 0.0281 moles of SCl4, we can create a ratio: \( n_{H2O} = 2 * n_{SCl4} = 2 * 0.0281\mathrm{~mol} = 0.0562\mathrm{~mol} \)
03

Convert moles of H2O to grams (part a)

Convert the moles of H2O needed to grams using its molar mass: \(m_{H2O} = n_{H2O} * M_{H2O} = 0.0562\mathrm{~mol} * 18.0\mathrm{~g/mol} = 1.0116\mathrm{~g}\) So, 1.0116 g of H2O will react with 5.000 g of SCl4.
04

Calculate moles of SO2 from H2O (part b)

The balanced chemical equation shows that 2 moles of H2O produce 1 mole of SO2. To find the moles of SO2 produced, we can create a ratio: \( n_{SO2} = \frac{1}{2} * n_{H2O}=\frac{1}{2} * 0.5556\mathrm{~mol} = 0.2778\mathrm{~mol} \)
05

Convert moles of SO2 to grams (part b)

Convert the moles of SO2 to grams using its molar mass: \(m_{SO2} = n_{SO2} * M_{SO2} = 0.2778\mathrm{~mol} * 64.1\mathrm{~g/mol} = 17.81\mathrm{~g}\) Therefore, 17.81 g of SO2 can be made from 10.00 g of H2O.
06

Calculate moles of HCl in the reaction (part c)

The balanced chemical equation shows that 1 mole of SCl4 produces 4 moles of HCl. So, 0.0281 moles of SCl4 will produce: \( n_{HCl} = 4 * n_{SCl4} = 4 * 0.0281\mathrm{~mol} = 0.1124\mathrm{~mol} \)
07

Convert moles of HCl to grams (part c)

Use HCl molar mass (M=36.5 g/mol) to convert moles to grams: \( m_{HCl} = n_{HCl} * M_{HCl} = 0.1124\mathrm{~mol} * 36.5\mathrm{~g/mol} = 4.1026\mathrm{~g} \) So, 4.1026 g of HCl will be formed when 5 g of SCl4 react with the calculated 1.0116 g of H2O.

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Most popular questions from this chapter

An organic compound of carbon and hydrogen has the empirical formula CH. What is its molecular formula if its molar mass is: (a) \(26 \mathrm{~g} / \mathrm{mol}\) (b) \(52 \mathrm{~g} / \mathrm{mol}\) (c) \(78 \mathrm{~g} / \mathrm{mol}\)

Tetraphosphorus decoxide, \(\mathrm{P}_{4} \mathrm{O}_{10}\), reacts with water to form phosphoric acid. If \(52.5 \mathrm{~g}\) of \(\mathrm{P}_{4} \mathrm{O}_{10}\) reacted with \(25.0 \mathrm{~g}\) of water, how many grams of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) could theoretically be produced?

Determine the empirical formula of the compound with the following mass percents of the elements present: \(66.63 \% \mathrm{C} ; 11.18 \% \mathrm{H} ; 22.19 \% \mathrm{O}\).

Can the actual yield ever be greater than the theoretical yield for a chemical reaction?

What is the empirical formula of a compound that is \(17.552 \% \mathrm{Na}, 39.696 \% \mathrm{Cr}\), and \(42.752 \% \mathrm{O} ?\)
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