Question

assuming that the coal was 2 90 % sulfur by mass and that combustion was complete, calculate the number of tons of sulfur dioxide produced by the plant during the year. express your answer numerically in tons. view available hint(s) tons so₂ submit previous answers incorrect; try again; 2 attempts remaining

Explanation:

1. First, establish the chemical - reaction equation for sulfur combustion:

• The chemical equation for the complete combustion of sulfur is \(S + O_2

ightarrow SO_2\). From this equation, the molar ratio of \(S\) to \(SO_2\) is \(1:1\).

• The molar mass of \(S\) is \(M_S = 32.06\ g/mol\), and the molar mass of \(SO_2\) is \(M_{SO_2}=32.06 + 2\times16.00=64.06\ g/mol\).

1. Let's assume the mass of coal burned in a year is \(m_{coal}\) tons. If the coal is \(2.90\%\) sulfur by mass, then the mass of sulfur in the coal is \(m_S = 0.029m_{coal}\) tons.

• Since the molar ratio of \(S\) to \(SO_2\) is \(1:1\), the mass of \(SO_2\) produced is related to the mass of \(S\) by the mass - ratio calculation.
• The mass of \(SO_2\) produced \(m_{SO_2}\) can be calculated using the ratio of their molar masses. Because the number of moles of \(S\) and \(SO_2\) is the same in the reaction, and \(n=\frac{m}{M}\) (where \(n\) is the number of moles, \(m\) is the mass, and \(M\) is the molar mass).
• We know that \(\frac{m_{SO_2}}{m_S}=\frac{M_{SO_2}}{M_S}\). Substituting \(M_S = 32.06\ g/mol\) and \(M_{SO_2}=64.06\ g/mol\), we get \(\frac{m_{SO_2}}{m_S}=\frac{64.06}{32.06}\approx2\).
• So, \(m_{SO_2}=2\times m_S\). Since \(m_S = 0.029m_{coal}\), then \(m_{SO_2}=2\times0.029m_{coal}=0.058m_{coal}\) tons.
• If we assume the mass of coal burned in a year is \(1\) ton (since the amount of coal is not given in the problem, and we are looking for the proportion), the mass of \(SO_2\) produced is \(m_{SO_2}=0.058\) tons when \(m_{coal} = 1\) ton.

Answer:

If the mass of coal burned is \(m_{coal}\) tons, the mass of \(SO_2\) produced is \(0.058m_{coal}\) tons. If we assume \(m_{coal}=1\) ton, the answer is \(0.058\) tons.