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(nh4)2cr207(s) - Cr2O3(s) + N2(g) + 4h20(g) 10.1 Calculate the number of molecules of Cr203 that is produced from the decomposition of 10 g of (nh4)

Accepted Answer

10 g of (NH4)2Cr2O7 contains 0.0412 moles of (NH4)2Cr2O7. The balanced equation shows that 1 mole of (NH4)2Cr2O7 produces 1 mole of Cr2O3. Therefore, 0.0412 moles of (NH4)2Cr2O7 will produce 0.0412 moles of Cr2O3.

Number of molecules = 0.0412 mol * 6.022 * 10^23 molecules/mol = 2.48 * 10^23 molecules

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