Question

170 lit of Sulphur dioxide will be produced at STP on complete burning of 250g of Sulphur true or false​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: state : \\ whether \: given \: statement \: \\ is \: true \: or \: false \: \\ \\ given \: that : \\ 170 \: l \: of \: Sulphur \: dioxide \: will \: be \\ produced \: at \: STP \: on \\ \: complete \: burning \: \: of \\ \: 250 \: g \: of \: Sulphur \\ \\ so \: we \: know \: that \\ S + O2 → SO2 \\ \\given \: \: mass \: of \: sulphur = 250 \: g \\ molar \: mass \: of \: sulpur = 3 2\: u \\ \\ we \: know \: that \\ number \: of \: moles = \frac{mass \: given}{molar \: mass} \\ \\ no \: of \: moles \: in \: S = \frac{250}{32} \\ \\ n \: (S)= \frac{125}{16}$

$but \: since \: here \\ stoichiometric \: \: coefficients \: of \\ all \: elements \: taking \: part \: in \: given \: \\ reaction \: is \: 1 \: only \\ \\ number \: of \: moles \: in \: SO2 = \frac{125}{16} \\ \\ so \: thus \: then \\ molar \: mass \: of \: SO2 = 32 + 2(16) \\ = 32 + 32 \\ = 64 \: u$

$now \: we \: know \: that \\ n = \frac{volume \: of \: gas \: (at \:STP) }{22.4} \\ \\ \frac{125}{16} = \frac{volume \: of \:SO2 }{22.4} \\ \\ = \frac{125}{16} \times 22.4 \\ \\ = 125 \times 1 .4 \\ \\ = 175.0 \\ \\volume \: of \: SO2 \: (at \:STP \: ) = 175 \: \: l$

$hence \: given \: statement \: is \: false$