How would I solve this?

If 1550 kJ of heat are produced by the reaction below, how many grams of MnCl2 (MW = 126 g/mol) are consumed?

MnCl2 + 2HNO3 → Mn(NO3)2 + 2HCl + 300 kJ

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  • Ash
    Lv 7
    1 month ago

    moles of MnCl2 consumed = (moles of MnCl2 from balanced equation / estimated heat) x actual heat produced

    moles of MnCl2 consumed = (1 mol / 300 kJ) x 1550 kJ = 5.17 mol

    mass of MnCl2 consumed = 5.17 mol x 126 g/mol = 651 g

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